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  • 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);
        }
    };
    

      

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