【LeetCode】98. 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:

Input:
    2
   / \
  1   3
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

二分查找树的中序遍历结果是一个递增序列，所以中序遍历这棵树，将节点值存储在一个vector里面，判断vector中的值是不是递增的，如果不是则返回false。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> tem;
    
    void helper(TreeNode* root){
        if(root==NULL) return;
        helper(root->left);
        tem.push_back(root->val);
        helper(root->right);
    }
    bool isValidBST(TreeNode* root) {
        if(root==NULL) return true;
        if(root->left==NULL && root->right==NULL) return true;
        helper(root);
        for(int i=1; i<tem.size(); i++)
            if(tem[i]<=tem[i-1]) return false;
        
        return true;
    }
};