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.
解法一:
利用二叉搜索树的特点,中序遍历然后判断是否严格升序。
/** * 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) { vector<int> ret; inOrder(root, ret); if(ret.size() <= 1) return true; for (int i = 1; i < ret.size(); i ++) { if(ret[i] <= ret[i-1]) return false; } return true; } void inOrder(TreeNode *root, vector<int>& ret) { if(!root) return; if(root->left) inOrder(root->left, ret); ret.push_back(root->val); if(root->right) inOrder(root->right, ret); } };
解法二:
由于只需要关心前一个节点与当前的大小关系,因此不用保存所有值。
只需要使用一个全局变量保存前一个节点值即可。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* last = NULL; bool isValidBST(TreeNode *root) { if(root == NULL) return true; else { bool leftRes = isValidBST(root->left); //short cut if(leftRes == false) return false; if(last && last->val >= root->val) return false; last = root; return isValidBST(root->right); } } };