Leetcode之Validate Binary Search Tree

题目：

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.

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 { private: vector<int> tranverse; public: void inorder(TreeNode* root) { if (root) { inorder(root->left); tranverse.push_back(root->val); inorder(root->right); } } bool isValidBST(TreeNode* root) { inorder(root); int len=tranverse.size(); for (int i=0; i < len-1; i++) { if (tranverse[i] >= tranverse[i + 1])return false; } return true; } };

想法：

   for循环第二个中最好不要放函数表达式