问题描述 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.
解决思路 利用递归，但每次需要传入一个判断区间。
代码
/** * 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; if (root->left == NULL && root->right == NULL) return true; long max = INT_MAX,min = INT_MIN; if (root->left == NULL) return root->right->val > root->val && helper(root->right,max,root->val,false,true); if (root->right == NULL) return root->left->val < root->val && helper(root->left,root->val,min,true,false); if (root->right->val <= root->val || root->left->val >= root->val) return false; return helper(root->right,max,root->val,false,true) && helper(root->left,root->val,min,true,false); } bool helper(TreeNode* root,int left_max, int right_min, bool left, bool right) { if (root == NULL) return true; if (!left&&(root->val > left_max || root->val <= right_min)) return false; if (!right&&(root->val >= left_max || root->val < right_min)) return false; if (left&&right&&(root->val >= left_max || root->val <= right_min)) return false; if (root->left == NULL && root->right == NULL) { return true; } if (root->left == NULL) return root->right->val > root->val && helper(root->right,left_max,root->val,left,true); if (root->right == NULL) return root->left->val < root->val && helper(root->left,root->val,right_min,true,right); if (root->right->val <= root->val || root->left->val >= root->val) return false; return helper(root->left,root->val,right_min,true,right) && helper(root->right,left_max,root->val,left,true); }};新闻热点
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