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