算法课第三周Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than the node’s key.
Both the left and right subtrees must also be binary search trees.
Example 1:
2
/ \
1 3
Binary tree [2,1,3], return true.
Example 2:
1
/ \
2 3
Binary tree [1,2,3], return false.
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【解析】
题意：判断一个二叉树是否为二分查找树。

何为二分查找树？1) 左子树的值都比根节点小；2) 右子树的值都比根节点大；3) 左右子树也必须满足上面两个条件。
需要注意的是，左子树的所有节点都要比根节点小，而非只是其左孩子比其小，右子树同样。这是很容易出错的一点是，很多人往往只考虑了每个根节点比其左孩子大比其右孩子小。我们对有序二叉树进行中序遍历得到的是一个有序的排列数组，从而可以进行判断。
代码如下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {     List<Integer> list = new ArrayList<Integer>();      public boolean isValidBST(TreeNode root) {          if (root == null) return true;          if (root.left == null && root.right == null) return true;          inOrderTraversal(root);          for (int i = 1; i < list.size(); i++) {              if (list.get(i) <= list.get(i - 1)) return false;          }          return true;       }      public void inOrderTraversal(TreeNode root) {          if (root == null) return;          inOrderTraversal(root.left);          list.add(root.val);          inOrderTraversal(root.right);      }}