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:

    2
   / 
  1   3

Input: [2,1,3]
Output: true

Example 2:

    5
   / 
  1   4
     / 
    3   6

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

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/**
 * 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:
    bool isValidBST(TreeNode* root) {
        return isValid(root,NULL,NULL);
    }
    
    bool isValid(TreeNode *root, int *l, int *r)
    {
        if(NULL==root)return true;
        if((l&&root->val<=*l)||(r&&root->val>=*r))return false;
        return isValid(root->left,l,&root->val) && isValid(root->right,&root->val,r);
    }
};

最简单的写法是递归, 但是注意c++这里不要用INT_MIN和INT_MAX, 因为有个test case会故意用这两个值来考导致fail. 用指针来绕过