AI代码解释 publicList<Integer>boundaryOfBinaryTree(TreeNode root){List<Integer>res=newArrayList<>();if(root==null)returnres;res.add(root.val);getBounds(root.left,res,true,false);getBounds(root.right,res,false,true);returnres;}publicvoidgetBounds(TreeNode node,List<Integer>res,boolean leftBoun...
The boundary of a binary tree is the concatenation of the root, the left boundary, the leaves ordered from left-to-right, and the reverse order of the right boundary. The left boundary is the set of nodes defined by the following: The root node's left child is in the left boundary. ...
Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root to the left-most node. Right boundary is defined ...
代码如下: publicList<Integer>boundaryOfBinaryTree(TreeNode root){ List<Integer> res =newArrayList<>();if(root ==null)returnres; res.add(root.val); getBounds(root.left, res,true,false); getBounds(root.right, res,false,true);returnres; }publicvoidgetBounds(TreeNode node, List<Integer> re...
publicList<Integer> boundaryOfBinaryTree(TreeNode root) { List<Integer> res = newArrayList<>(); if(root == null) returnres; res.add(root.val); getBounds(root.left, res, true, false); getBounds(root.right, res, false, true); returnres; } publicvoidgetBounds(TreeNode node, List<Inte...
祖传的手艺不想丢了,所以按顺序写一个leetcode的题解。计划每日两题,争取不卡题吧 常规题号的题目已经写到头了,开始从头补vip题的题解 545.二叉树的边界 力扣leetcode.cn/problems/boundary-of-binary-tree/ 根据题目的描述,二叉树的边界由根节点、左边界、逆序右边界以及叶节点组成。在这里需要得到从上向下...
classSolution{vector<int>ans;public:vector<int>boundaryOfBinaryTree(TreeNode*root){if(!root)return{};ans.push_back(root->val);dfs(root->left,-1);dfs(root->right,1);returnans;}voiddfs(TreeNode*root,int dir){if(!root)return;if(dir==-1)//左边界{ans.push_back(root->val);//前序...
标签:树、二叉树 解法一: class Solution: def boundaryOfBinaryTree(self, root: TreeNode) -> List[int]: if not root: return [] # 寻找左边界 left = [] node = root if node.left: while node: left.append(node) if node.left:
545. Boundary of Binary Tree 题目:Recursion 给定一个树形结构,返回 [ 根节点,左边界(为一个list),从左到右的所有叶子结点,及反序右边界] 的concatenation 左边界中的节点包括: 根节点的左儿子。若根节点无左儿子,左边界中无节点 左边界中节点的左儿子,或无左儿子的节点的右儿子 须删掉最左的叶节点。* 由...
545 Boundary of Binary Tree Algorithms Medium 544 Output Contest Matches Algorithms Medium 543 Diameter of Binary Tree Algorithms Easy 542 01 Matrix Algorithms Medium 541 Reverse String II Algorithms Easy 540 Single Element in a Sorted Array Draft Medium 539 Minimum Time Difference Algorithms Medium 53...