commit

Author: Wang-Jun-Chao

[096][Unique Binary Search Trees]/src/Solution.java

@@ -7,42 +7,40 @@
 public class Solution {
     /**
      * Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
-     *
+     * <p>
      * For example,
      * Given n = 3, there are a total of 5 unique BST's.
-     *  1         3     3      2      1
-     *   \       /     /      / \      \
-     *   3     2     1       1   3      2
-     *  /     /       \                 \
+     * 1         3     3      2      1
+     * \       /     /      / \      \
+     * 3     2     1       1   3      2
+     * /     /       \                 \
      * 2     1         2                 3
-     *
+     * <p>
      * µÝÍƹ«Ê½
-     * f(k)*f(n-1-k)：f(k)表示根结点左子树有k个结点，其有的形状是f(k)，f(n-1-k)表示右子树有n-1-k个结点
-     *
-     * f(n) = 2*f(n-1) + f(1)*f(n-2) + f(2)f(n-3) + f(3)f(n-4) + ... +f(n-2)*f(1)
+     * f(0) = 1
+     * f(1) = 1
+     * f(i) = f(0)f(i-1) + f(1)f(i-1) + ... + f(i-1)f(0)
      *
      * @param n
      * @return
      */
     public int numTrees(int n) {
 
         if (n <= 0) {
-            return 0;
+            return 1;
         } else if (n == 1) {
             return 1;
         }
 
         int[] result = new int[n + 1];
-        result[0] = 0;
+        result[0] = 1;
         result[1] = 1;
 
 
         // Çóf(2)...f(n)
         for (int i = 2; i <= n; i++) {
-            // 求f(i)
-            result[i] = 2 * result[i - 1];
-            for (int j = 1; j <= i - 1 ; j++) {
-                result[i] += result[j]*result[i - 1 -j];
+            for (int j = 1; j <= i; j++) {
+                result[i] += result[j - 1] * result[i - j];
             }
 
         }

[096][Unique Binary Search Trees]/src/Solution2.java

@@ -0,0 +1,51 @@
+/**
+ * Author: 王俊超
+ * Date: 2015-06-18
+ * Time: 17:36
+ * Declaration: All Rights Reserved !!!
+ */
+public class Solution2 {
+    /**
+     * 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.
+     *  1         3     3      2      1
+     *   \       /     /      / \      \
+     *   3     2     1       1   3      2
+     *  /     /       \                 \
+     * 2     1         2                 3
+     *
+     * 递推公式
+     * f(k)*f(n-1-k)：f(k)表示根结点左子树有k个结点，其有的形状是f(k)，f(n-1-k)表示右子树有n-1-k个结点
+     *
+     * f(n) = 2*f(n-1) + f(1)*f(n-2) + f(2)f(n-3) + f(3)f(n-4) + ... +f(n-2)*f(1)
+     *
+     * @param n
+     * @return
+     */
+    public int numTrees(int n) {
+
+        if (n <= 0) {
+            return 1;
+        } else if (n == 1) {
+            return 1;
+        }
+
+        int[] result = new int[n + 1];
+        result[0] = 0;
+        result[1] = 1;
+
+
+        // 求f(2)...f(n)
+        for (int i = 2; i <= n; i++) {
+            // 求f(i)
+            result[i] = 2 * result[i - 1];
+            for (int j = 1; j <= i - 1 ; j++) {
+                result[i] += result[j]*result[i - 1 -j];
+            }
+
+        }
+        return result[n];
+    }
+}

[106][Construct Binary Tree From Inorder And Postorder Traversal]/src/Solution.java

@@ -31,8 +31,7 @@ public class Solution {
     public TreeNode buildTree(int[] inorder, int[] postorder) {
 
         // 参数检验
-        if (inorder == null || postorder == null || inorder.length == 0
-                || inorder.length != postorder.length) {
+        if (inorder == null || postorder == null || inorder.length == 0 || inorder.length != postorder.length) {
             return null;
         }
 
@@ -70,13 +69,15 @@ else if (x < y) {
                 }
 
                 // 左子树非空，构建左子树
+                // [x, x+1, ..., idx-1] -> 总计 idx-x个
                 int leftLength = idx - x;
                 if (leftLength > 0) {
                     // i, i + leftLength - 1，前序遍历的左子树的起始，结束位置
                     root.left = solve(inorder, x, idx - 1, postorder, i, i + leftLength - 1);
                 }
 
                 // 右子树非空，构建右子树
+                // [idx+1, idx+2, ..., y] -> 总计 y-idx个
                 int rightLength = y - idx;
                 if (rightLength > 0) {
                     // i + leftLength, j - 1，前序遍历的右子树的起始，结束位置