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);
    }
};