leetcode Validate Binary Search Tree
2013-11-27 20:44
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Validate Binary Search Tree
Total Accepted: 3218 TotalSubmissions: 12801My Submissions
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}".
Take notice that
The left subtree of a node contains only nodes with keys less
than the node's key
I.e., all nodes are less than the root.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool check(TreeNode *root, int min, int max) { if (!root) return true; if (root->val <= min || root->val >= max) return false; if (root->left && !check(root->left, min, root->val)) return false; if (root->right && !check(root->right, root->val, max)) return false; return true; } bool isValidBST(TreeNode *root) { if (!root) return true; return check(root, INT_MIN, INT_MAX); } };
Another method is to give an inorder traversal of Binary Tree. For example,
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isValidBST(TreeNode *root) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. //注意题目要求是 less than和greater than; stack<TreeNode*> S; TreeNode *pre = NULL, *p = root; while(p || S.empty() == false) { while(p) { S.push(p); p = p->left; } if(S.empty() == false) { p = S.top(); S.pop(); if(pre && p->val <= pre->val)return false; pre = p; p = p->right; } } return true; } };
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