LeetCode 98. Validate Binary Search Tree--C++解法--判断是否是BST--递归，迭代做法，中序遍历

最新推荐文章于 2021-01-29 15:56:23 发布

zhangpeterx

最新推荐文章于 2021-01-29 15:56:23 发布

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分类专栏： LeetCode c++-做题

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LeetCode 98. Validate Binary Search Tree–C++解法–判断是否是BST–递归，迭代做法，中序遍历

LeetCode题解专栏：LeetCode题解
LeetCode 所有题目总结：LeetCode 所有题目总结
大部分题目C++，Python，Java的解法都有。

题目地址：Validate Binary Search Tree - LeetCode

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.

这道题目是判断二叉树是否是搜索二叉树，自然就想到了递归和迭代2种做法。

递归的话在考虑后发现对每个节点都要维护一个最小值和最大值。
C++解法如下：

/**
 * 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