You are given therootof a binary search tree (BST) and an integerval. Find the node in the BST that the node's value equalsvaland return the subtree rooted with that node. If such a node does not exist, returnn
将所有节点值依次入队列,在入队列前先判断节点值是否等于val,等于就直接返回当前节点所在子树。 publicTreeNode searchBST(TreeNode root, intval) {if(root ==null|| root.val==val) {returnroot; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root);while(!queue.isEmpty()) { ...
Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL. For example, Given the tree: 4 / \ ...
Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL. 例如, 代码语言:javascript 代码运行次数:...
Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node’s value equals the given value. Return the subtree rooted with that node. If such node doesn’t exist, you should return NULL. ...
题目地址:https://leetcode.com/problems/search-in-a-binary-search-tree/description/ 题目描述 Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node’s value equals the given value. Return the subtree rooted with...
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Right } } // 没有找到值为 val 的结点,则返回空 return nil } 题目链接: Search in a Binary Search Tree : leetcode.com/problems/s 二叉搜索树中的搜索: leetcode-cn.com/problem LeetCode 日更第 89 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
# 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 BSTIterator: def __init__(self, root: Optional[TreeNode]): # 初始化一个空结点栈(所有结点的左子结点...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicTreeNodelowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q){if(root.val-p.val>0&&root.val-q.val...