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.
confused what "{1,#,2,3}"
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read more on how binary tree is serialized on OJ.
方法一:根据二叉查找树中序遍历是递增序列的特点,在进行中序遍历时,如果发现某个节点的值比它的先序节点要大,那么就是非法的。
class Solution { public: bool isValidBST(TreeNode* root,TreeNode*& pre) { if(!root)return true; bool flag = isValidBST(root->left,pre); if(!flag)return false; if(pre && pre->val >= root->val)return false; pre = root; return isValidBST(root->right,pre); } bool isValidBST(TreeNode* root) { TreeNode* pre = NULL; return isValidBST(root,pre); } };
方法二:根据二叉查找树的定义,每一节点都在左右子树值的中间,因此,每次递归遍历时,给定一个值的范围,该范围是当前子树值的两个边界,用该范围来判断合法性,该方法的效率比方法一高
struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { public: bool isValidBST(TreeNode* root,int minValue,int maxValue)//左右子树的范围 { if(!root)return true; if(minValue < root->val && root -> val < maxValue) { return isValidBST(root->left,minValue,root->val) && isValidBST(root->right,root->val,maxValue); } return false; } bool isValidBST(TreeNode* root) { return isValidBST(root,INT_MIN,INT_MAX); } };
时间: 2024-11-06 03:55:00