98. Validate Binary Search Tree

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.
解题思路：

该题是让判断一棵二叉树是否是一颗二分搜索树，即：左小于根，根小于右。故可以用dfs来给出中序排序顺序，再判断是否递增即可。

class Solution {

public:

    bool isValidBST(TreeNode* root) {

        if(root==NULL) return true;

        vector<int> nums;

        inOrder(root,nums);

        for(int i=0;i<nums.size()-1;i++){

            if(nums[i]>=nums[i+1])

            return false;

        }

        return true;

    }

    void inOrder(TreeNode* root,vector<int> &num)

    {

        if(root==NULL) return;

        inOrder(root->left,num);

        num.push_back(root->val);

        inOrder(root->right,num);

    }

};