98 Validate Binary Search Tree

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.

/** * 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 checkValidBST(root, LONG_MIN, LONG_MAX);    }        bool checkValidBST(TreeNode* root, long leftVal, long rightVal) {    if(!root) return true;    int curVal = root->val;    if(curVal <=leftVal || curVal >= rightVal)        return false;    return checkValidBST(root->left, leftVal, curVal) && checkValidBST(root->right, curVal, rightVal);}};

判断一棵树是否是二分搜索树：

1、左子树的值都比根小

2、右子树的值都比根大

3、左右子树也必须满足上述条件。

因而递归先序遍历二叉树看看每个节点是否满足上述条件即可