【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.

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

题意：判断一棵树是否是二叉搜索树；

思路：中序遍历看是否是有序的

class Solution{public:bool isValidBST(TreeNode* root){int pre = INT_MIN;bool res = true;inorder(pre, root, res);return res;}void inorder(int& pre, TreeNode* root,  bool &res){if (root){inorder(pre, root->left, res);if (root->val<=pre){res = false;return;}pre = root->val;inorder(pre, root->right, res);}}};