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.

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}".
 
用两个值分别作max和min不是一个好办法。
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode *root) {
        if(root == NULL)return 1;
        return isBST(root->left,-1000000,root->val) && isBST(root->right,root->val,1000000);
    }
    bool isBST(TreeNode *root, int min ,int max )   
    {
        if(root == NULL)return 1;
        if(root->val >=max || root->val <= min)return 0;
        return isBST(root->left,min,root->val) && isBST(root->right,root->val,max);
    }
};

  

原文地址:https://www.cnblogs.com/pengyu2003/p/3606315.html