leetcode-98-Validate Binary Search Tree
98. Validate Binary Search Tree
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- Total Submissions: 626297
- Difficulty: Medium
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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 3Binary tree
[2,1,3]
, return true.
Example 2:
1 / 2 3Binary tree
[1,2,3]
, return false.
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(1), 使用递归解决。
(2), 没想到测试用例这么极端, 居然就一定要到 int 的范围尽头, 记住 int 的范围 [-2147483648, 2147483647]
/** * 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 Judge(TreeNode* root, int leftval, int rightval){ if(root == NULL){ return true; } if(root->val < leftval || root->val > rightval){ return false; } if(root->left && root->right){ if(root->left->val < root->val && root->right->val > root->val){ return Judge(root->left, leftval, root->val-1) && Judge(root->right, root->val+1, rightval); }else{ return false; } }else if(root->left != NULL){ if(root->left->val < root->val){ return Judge(root->left, leftval, root->val-1); }else{ return false; } }else if(root->right != NULL){ if(root->right->val > root->val){ return Judge(root->right, root->val+1, rightval); }else{ return false; } }else{ return true; } } bool isValidBST(TreeNode* root) { return Judge(root, -2147483648, 2147483647); } };