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}67boolcheckValid(TreeNode * root,intleft,intright)8{9if(root ==NULL)10returntrue;11return(root->...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {private:intpre; vector<int>vec;public: Solution():pre(LONG_MIN) {}boolisValidBST(TreeNode...
return dfs(root.Left, low, &root.Val) && dfs(root.Right, &root.Val, high) } 题目链接: Validate Binary Search Tree: leetcode.com/problems/v 带因子的二叉树: leetcode.cn/problems/va LeetCode 日更第 209 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 ...
这段code 的最大值最小值不应是2147483647 和-2147483648。应该选稍微小一点的下界和稍微大一点的上界,否则[2147483647]这个case过不了,为了方便可以取10**10>2147483647和-10**1<-2147483648, 参考http://chaoren.is-programmer.com/posts/42736.html class Solution: # @param root, a tree node # @return ...
Leetcode: Validate Binary Search Tree,需要保证的是左子树所有节点都小于根,而并非仅仅左子树根。所以Recursion里面要存的是可以取值的范围。其实就是对于每个结点保存左右界,也就是保证结点满足它的左子树的每个结点比当前结点值小,右子树的每个结点比当前结点值大。
LeetCode 98 Validate Binary Search Tree LeetCode 98 Validate Binary Search Tree Given a binary tr... ShuiLocked阅读 281评论 0赞 0 JS中的算法与数据结构——二叉查找树(Binary Sort Tree) 二叉查找树(Binary Sort Tree) 我们之前所学到的列表,栈等都是一种线性的数据结构,今天我们将学习计... Cryptic...
(TreeNode*root){returnjudge(root,INT_MIN,INT_MAX);}booljudge(TreeNode*root,intlow,inthigh){if(root==NULL)returntrue;returnroot->val>low&&root->val<high&&judge(root->left,low,root->val)&&judge(root->right,root->val,high);}};
98. Validate Binary Search Tree #38 Open hayashi-ay wants to merge 4 commits into main from hayashi-ay-patch-27 +200 −0 Conversation 1 Commits 4 Checks 0 Files changed 1 ConversationOwner hayashi-ay commented Feb 29, 2024 https://leetcode.com/problems/validate-binary-search-tree/...
Leetcode-Medium 98. Validate Binary Search Tree 题目描述 判定一棵树是否满足二叉搜索树的性质。二叉查找树(Binary Search Tree),(又:二叉搜索树,二叉排序树)它或者是一棵空树,或者是具有下列性质的二叉树: 若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值; 若它的右子树不空,则右子树上所有...
public class Solution { TreeNode pre = null; public boolean isValidBST(TreeNode root) { return dfs(root); } public boolean dfs(TreeNode root) { if (root == null) return true; if (!dfs(root.left)) return false; if (pre != null && pre.val >= root.val) return false; ...