# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def isValidBST(self, root: Optional[TreeNode]) -> bool: return Solution.dfs(root, None, Non...
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. 一开始是这样做的: 1classSolution {2public:3boolisValidBST(TreeNode*root) {4returncheckValid(root, INT_MIN, INT_MAX);5}67boolcheck...
1 class Solution { 2 public: 3 bool isValidBST(TreeNode* root)//二叉树的中序遍历 4 { 5 //递归出口 6 if(!root) return true; 7 //搜索二叉树中根结点应小于左子树的最右边节点(左子树最大值),大于右子树的最左边节点(右子树最小值) 8 bool root_valid = false; 9 TreeNode* most_left_...
publicclassSolution {publicbooleanisValidBST(TreeNode root) {if(root ==null)returntrue;returnvalid(root, Long.MIN_VALUE, Long.MAX_VALUE); }publicbooleanvalid(TreeNode root,longlow,longhigh) {if(root ==null)returntrue;if(root.val <= low || root.val >= high)returnfalse;returnvalid(root....
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...
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....
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{publicbooleanisValidBST(TreeNoderoot){Integermin=null;Integermax=null;returnisValid(root,min,max);}publicbooleanisValid...
validate(root.right) if right_min <= root.val: self.flag = False return min(left_min, right_min, root.val), max(left_max, right_max, root.val) def isValidBST(self, root: TreeNode) -> bool: if root is None: return True self.flag = True self.validate(root) return self.flag...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{longpre=(long)Integer.MIN_VALUE-1;publicbooleanisValidBST(TreeNoderoot){if(root==null){returntrue;}if(isValid...
我写了一个很漂亮的递归检查左节点小右节点大，然后被5，4，6，n，n，3，7打败了 中