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) {if (root == NULL) return true;long long pre = LLONG_MIN;///bool valid = true;dfs(root, pre,valid);return valid;}private:void dfs(TreeNode* root, long long& pre, bool& valid){if (root->left)dfs(root->left, pre,valid);if (root->val <= pre){valid = false;return;}elsepre = root->val;if (valid&&root->right)dfs(root->right, pre,valid);}};

更简洁的方法

class Solution {public:    bool isValidBST(TreeNode* root) {        TreeNode* prev = NULL;        return validate(root, prev);    }    bool validate(TreeNode* node, TreeNode* &prev) {        if (node == NULL) return true;        if (!validate(node->left, prev)) return false;        if (prev != NULL && prev->val >= node->val) return false;        prev = node;        return validate(node->right, prev);    }};