Validate Binary Search Tree

Validate Binary Search Tree

Total Accepted: 73179 Total Submissions: 359616 Difficulty: Medium

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}".

 

BST的中序遍历结果是有序的，可根据这个性质进行判读

 

/**
 * 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 inorder(TreeNode* root,TreeNode** pre){
        if(!root) return true;
        
        bool is_left_valid = inorder(root->left,pre);
        
        if(!is_left_valid) return false;
        
        if(*pre != NULL){
            if((*pre)->val >= root->val){
                return false;
            }
        }
        
        *pre =root;
        
        return inorder(root->right,pre);
    }
    bool isValidBST(TreeNode* root) {
        TreeNode* pre = NULL;
        return inorder(root,&pre);
    }
};
//[1,1]
//[5,3,8,2,4,7,9]
//[10,5,15,null,null,6,20]