fd

Author: liwentian

ebook/tree/Tree.aux

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ebook/tree/Tree.tex

@@ -78,4 +78,207 @@ \subsubsection{Solution II}
 
     return root;
 }
-\end{Code}
+\end{Code}
+
+\newpage
+
+\section{Same Tree} %%%%%%%%%%%%%%%%%%%%%%
+
+\subsubsection{Description}
+
+Given two binary trees, write a function to check if they are equal or not.
+
+Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
+
+\subsubsection{Solution}
+
+\begin{Code}
+public boolean isSameTree(TreeNode p, TreeNode q) {
+    if (p == null && q == null) {
+        return true;
+    }
+    if (p == null || q == null) {
+        return false;
+    }
+    return p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+}
+\end{Code}
+
+\newpage
+
+\section{Binary Search Tree Iterator} %%%%%%%%%%%%%%%%%%%%%%
+
+\subsubsection{Description}
+
+Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
+
+Calling next() will return the next smallest number in the BST.
+
+\textbf{Note:} next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
+
+\subsubsection{Solution}
+
+\begin{Code}
+public class BSTIterator {
+
+    private Stack<TreeNode> mStack;
+    private TreeNode mCurNode;
+
+    public BSTIterator(TreeNode root) {
+        mStack = new Stack<TreeNode>();
+        mCurNode = root;
+    }
+
+    public boolean hasNext() {
+        return !mStack.isEmpty() || mCurNode != null;
+    }
+
+    public int next() {
+        int result = -1;
+
+        while (hasNext()) {
+            if (mCurNode != null) {
+                mStack.push(mCurNode);
+                mCurNode = mCurNode.left;
+            } else {
+                mCurNode = mStack.pop();
+                result = mCurNode.val;
+                mCurNode = mCurNode.right;
+                break;
+            }
+        }
+
+        return result;
+    }
+}
+\end{Code}
+
+\newpage
+
+\section{Unique Binary Search Trees} %%%%%%%%%%%%%%%%%%%%%%
+
+\subsubsection{Description}
+
+Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
+
+For example,
+
+Given n = 3, there are a total of 5 unique BST's.
+\begin{Code}
+   1         3     3      2      1
+    \       /     /      / \      \
+     3     2     1      1   3      2
+    /     /       \                 \
+   2     1         2                 3
+\end{Code}
+
+\subsubsection{Solution}
+\begin{Code}
+public int numTrees(int n) {
+    int[] dp = new int[n + 1];
+    dp[0] = 1;
+
+    for (int i = 1; i <= n; i++) {
+        for (int j = 0; j < i; j++) {
+            dp[i] += dp[j] * dp[i - j - 1];
+        }
+    }
+
+    return dp[n];
+}
+\end{Code}
+
+\newpage
+
+\section{Lowest Common Ancestor of a Binary Search Tree} %%%%%%%%%%%%%%%%%%%%%%
+
+\subsubsection{Description}
+
+Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
+
+According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
+\begin{Code}
+        _______6______
+       /              \
+    ___2__          ___8__
+   /      \        /      \
+   0      _4       7       9
+         /  \
+         3   5
+\end{Code}
+For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
+
+\subsubsection{Solution}
+
+\begin{Code}
+// 耗时9ms
+public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+    if (!checkExist(root, p) || !checkExist(root, q)) {
+        throw new IllegalArgumentException("Not exist!!");
+    }
+
+    while (root != null) {
+        if (p.val < root.val && q.val < root.val) {
+            root = root.left;
+        } else if (p.val > root.val && q.val > root.val) {
+            root = root.right;
+        } else {
+            break;
+        }
+    }
+    return root;
+}
+
+/**
+ * 如何判断p或q一定存在，如果是BST就很简单
+ */
+private boolean checkExist(TreeNode root, TreeNode node) {
+    TreeNode cur = root;
+    while (cur != null) {
+        if (node.val > cur.val) {
+            cur = cur.right;
+        } else if (node.val < cur.val) {
+            cur = cur.left;
+        } else {
+            return true;
+        }
+    }
+    return false;
+}
+\end{Code}
+
+\newpage
+
+\section{Balanced Binary Tree} %%%%%%%%%%%%%%%%%%%%%%
+
+\subsubsection{Description}
+
+Given a binary tree, determine if it is height-balanced.
+
+For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
+
+\subsubsection{Solution}
+
+\begin{Code}
+public boolean isBalanced(TreeNode root) {
+    return isBalanced(root, null);
+}
+
+private boolean isBalanced(TreeNode root, int[] height) {
+    if (root == null) {
+        return true;
+    }
+
+    int[] left = new int[1], right = new int[1];
+
+    boolean result = isBalanced(root.left, left) && isBalanced(root.right, right);
+
+    if (height != null) {
+        height[0] = Math.max(left[0], right[0]) + 1;
+    }
+
+    return result && Math.abs(left[0] - right[0]) <= 1;
+}
+\end{Code}
+
+\newpage

ebook/tree/leetcode-tree.log

@@ -1,4 +1,4 @@
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-Output written on leetcode-tree.pdf (4 pages).
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ebook/tree/leetcode-tree.pdf

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