二叉搜索树的判定

题目描述

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.

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

二分查找树的中序遍历结果是一个递增序列。

实现代码：

    

 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };

class Solution {
public:
    vector<int> result;
    bool isValidBST(TreeNode *root) {
        if(root==NULL) return true;
        if(root->left==NULL && root->right==NULL) return true;
        inorder(root);
        int size=result.size();
        for(int i=1;i<size;i++)
        {
            if(result[i]<=result[i-1])
                return false;
        }
        return true;
    }
    void inorder(TreeNode*root)
    {
        if(root==NULL) return;
        inorder(root->left);
        result.push_back(root->val);
        inorder(root->right);
    }
};