Validate Binary Search Tree

最新推荐文章于 2025-08-12 12:49:47 发布

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#C++ #leetcode #二叉树

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本文介绍了一种判断给定二叉树是否为有效二叉搜索树的方法。通过递归遍历，确保每个节点的左子树的所有值小于该节点值，而右子树的所有值大于该节点值。

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一、问题描述

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

Binary tree [2,1,3], return true.

Example 2:

    1
   / \
  2   3

Binary tree [1,2,3], return false.

二、思路

按照二叉树的定义，对一个节点来说，左子树所有值都小于它，右子树所有值都大于它。

三、代码

/**
 * 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){
        TreeNode* pre = NULL;
        return isValid(root,pre);
    }
    bool isValid(TreeNode* root,TreeNode* &prev){
        if(root == NULL) return true;
        if (!isValid(root->left, prev)) return false;
        if (prev != NULL && prev->val >= root->val) return false;
        prev = root;
        return isValid(root -> right, prev);
    }
};