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 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   3Input: [2,1,3]
Output: true

Example 2:

    5/ \1   4/ \3   6Input: [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(TreeNode *root) {return isValidBST(root, nullptr, nullptr);}bool isValidBST(TreeNode *root, TreeNode *min, TreeNode *max) {if (!root) return true;if (min != nullptr && root->val <= min->val) return false;if (max != nullptr && root->val >= max->val) return false;return isValidBST(root->left, min, root) && isValidBST(root->right, root, max);}
};

可以采用中序遍历，看是否有序。

/*** 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) {if (root == nullptr) return true;stack<TreeNode*> s;TreeNode* current = root; TreeNode* prev = nullptr;while (current || !s.empty()){while(current){s.push(current);current = current->left;}current = s.top(); s.pop();if (prev && prev->val >= current->val) return false;prev = current;current = current->right;} return true;}
};

奇怪的是中序遍历花费的时间跟递归做法花费的时间基本一样。

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