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Tree & 20230506 daily chanllenge
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‎Tree/104.maximum-depth-of-binary-tree.cpp

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/*
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* @lc app=leetcode id=104 lang=cpp
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*
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* [104] Maximum Depth of Binary Tree
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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int maxDepth(TreeNode* root) {
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if(root==NULL) return 0;
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int lh=maxDepth(root->left);
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int rh=maxDepth(root->right);
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return 1+max(lh,rh);
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}
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};
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// @lc code=end
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‎Tree/105.construct-binary-tree-from-preorder-and-inorder-traversal.cpp

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/*
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* @lc app=leetcode id=105 lang=cpp
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*
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* [105] Construct Binary Tree from Preorder and Inorder Traversal
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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unordered_map<int, int> valToIndex;
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public:
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TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
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for (int i = 0; i < inorder.size(); i++) {
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valToIndex[inorder[i]] = i;
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}
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return build(preorder, 0, preorder.size() - 1,
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inorder, 0, inorder.size() - 1);
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}
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TreeNode* build(vector<int>& preorder, int preStart, int preEnd,
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vector<int>& inorder, int inStart, int inEnd) {
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if (preStart > preEnd) {
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return nullptr;
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}
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// root 节点对应的值就是前序遍历数组的第一个元素
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int rootVal = preorder[preStart];
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// rootVal 在中序遍历数组中的索引
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int index = valToIndex[rootVal];
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int leftSize = index - inStart;
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// 先构造出当前根节点
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TreeNode* root = new TreeNode(rootVal);
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// 递归构造左右子树
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root->left = build(preorder, preStart + 1, preStart + leftSize,
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inorder, inStart, index - 1);
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root->right = build(preorder, preStart + leftSize + 1, preEnd,
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inorder, index + 1, inEnd);
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return root;
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}
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};
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// @lc code=end
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‎Tree/114.flatten-binary-tree-to-linked-list.cpp

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/*
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* @lc app=leetcode id=114 lang=cpp
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*
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* [114] Flatten Binary Tree to Linked List
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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void flatten(TreeNode* root) {
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// base case
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if (root == nullptr) return;
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// 利用定义，把左右子树拉平
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flatten(root->left);
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flatten(root->right);
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/**** 后序遍历位置 ****/
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// 1、左右子树已经被拉平成一条链表
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TreeNode* left = root->left;
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TreeNode* right = root->right;
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// 2、将左子树作为右子树
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root->left = nullptr;
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root->right = left;
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// 3、将原先的右子树接到当前右子树的末端
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TreeNode* p = root;
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while (p->right != nullptr) {
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p = p->right;
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}
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p->right = right;
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}
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};
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// @lc code=end
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‎Tree/116.populating-next-right-pointers-in-each-node.cpp

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/*
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* @lc app=leetcode id=116 lang=cpp
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*
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* [116] Populating Next Right Pointers in Each Node
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*/
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// @lc code=start
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/*
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// Definition for a Node.
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class Node {
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public:
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int val;
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Node* left;
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Node* right;
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Node* next;
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Node() : val(0), left(NULL), right(NULL), next(NULL) {}
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Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
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Node(int _val, Node* _left, Node* _right, Node* _next)
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: val(_val), left(_left), right(_right), next(_next) {}
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};
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*/
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class Solution {
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public:
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Node* connect(Node* root) {
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if (root == nullptr) return nullptr;
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// 遍历「三叉树」，连接相邻节点
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traverse(root->left, root->right);
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return root;
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}
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// 三叉树遍历框架
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void traverse(Node* node1, Node* node2) {
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if (node1 == nullptr || node2 == nullptr) {
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return;
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}
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/**** 前序位置 ****/
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// 将传入的两个节点穿起来
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node1->next = node2;
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// 连接相同父节点的两个子节点
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traverse(node1->left, node1->right);
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traverse(node2->left, node2->right);
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// 连接跨越父节点的两个子节点
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traverse(node1->right, node2->left);
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}
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};
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// @lc code=end
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‎Tree/144.binary-tree-preorder-traversal.cpp

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/*
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* @lc app=leetcode id=144 lang=cpp
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*
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* [144] Binary Tree Preorder Traversal
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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vector<int> ans;
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public:
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vector<int> preorderTraversal(TreeNode* root) {
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if (root) {
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ans.push_back(root->val);
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preorderTraversal(root->left);
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preorderTraversal(root->right);
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}
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return ans;
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}
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};
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// @lc code=end
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‎Tree/226.invert-binary-tree.cpp

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/*
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* @lc app=leetcode id=226 lang=cpp
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*
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* [226] Invert Binary Tree
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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TreeNode* invertTree(TreeNode* root) {
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if (root == nullptr) {
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return nullptr;
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}
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// 利用函数定义，先翻转左右子树
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TreeNode* left = invertTree(root->left);
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TreeNode* right = invertTree(root->right);
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// 然后交换左右子节点
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root->left = right;
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root->right = left;
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// 和定义逻辑自恰：以 root 为根的这棵二叉树已经被翻转，返回 root
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return root;
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}
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};
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// @lc code=end
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‎Tree/297.serialize-and-deserialize-binary-tree.cpp

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/*
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* @lc app=leetcode id=297 lang=cpp
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*
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* [297] Serialize and Deserialize Binary Tree
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Codec {
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public:
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string serialize(TreeNode* root) {
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ostringstream out;
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serialize(root, out);
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return out.str();
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}
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TreeNode* deserialize(string data) {
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istringstream in(data);
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return deserialize(in);
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}
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private:
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void serialize(TreeNode* root, ostringstream& out) {
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if (root) {
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out << root->val << ' ';
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serialize(root->left, out);
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serialize(root->right, out);
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} else {
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out << "# ";
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}
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}
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TreeNode* deserialize(istringstream& in) {
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string val;
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in >> val;
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if (val == "#")
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return nullptr;
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TreeNode* root = new TreeNode(stoi(val));
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root->left = deserialize(in);
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root->right = deserialize(in);
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return root;
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}
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};
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// Your Codec object will be instantiated and called as such:
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// Codec ser, deser;
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// TreeNode* ans = deser.deserialize(ser.serialize(root));
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// @lc code=end
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‎Tree/315.count-of-smaller-numbers-after-self.cpp

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/*
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* @lc app=leetcode id=315 lang=cpp
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*
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* [315] Count of Smaller Numbers After Self
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*/
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// @lc code=start
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class Solution {
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public:
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struct Pair {
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int val, id;
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Pair(int val, int id) {
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// 记录数组的元素值
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this->val = val;
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// 记录元素在数组中的原始索引
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this->id = id;
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}
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};
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// 归并排序所用的辅助数组
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Pair* temp;
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// 记录每个元素后面比自己小的元素个数
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int* count;
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// 主函数
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vector<int> countSmaller(vector<int>& nums) {
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int n = nums.size();
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count = new int[n]();
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temp = new Pair[n]();
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Pair* arr = new Pair[n];
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// 记录元素原始的索引位置，以便在 count 数组中更新结果
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for (int i = 0; i < n; i++)
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arr[i] = Pair(nums[i], i);
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// 执行归并排序，本题结果被记录在 count 数组中
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sort(arr, 0, n - 1);
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vector<int> res;
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for (int i = 0; i < n; i++)
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res.push_back(count[i]);
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delete[] count;
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delete[] temp;
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return res;
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}
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// 归并排序
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void sort(Pair* arr, int lo, int hi) {
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if (lo == hi) return;
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int mid = lo + (hi - lo) / 2;
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sort(arr, lo, mid);
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sort(arr, mid + 1, hi);
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merge(arr, lo, mid, hi);
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}
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// 合并两个有序数组
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void merge(Pair* arr, int lo, int mid, int hi) {
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for (int i = lo; i <= hi; i++) {
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temp[i] = arr[i];
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}
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int i = lo, j = mid + 1;
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for (int p = lo; p <= hi; p++) {
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if (i == mid + 1) {
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arr[p] = temp[j++];
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} else if (j == hi + 1) {
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arr[p] = temp[i++];
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// 更新 count 数组
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count[arr[p].id] += j - mid - 1;
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} else if (temp[i].val > temp[j].val) {
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arr[p] = temp[j++];
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} else {
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arr[p] = temp[i++];
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// 更新 count 数组
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count[arr[p].id] += j - mid - 1;
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}
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}
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}
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};
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// @lc code=end
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