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);
}
}
};版权声明：本文为lplp90908原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。