Leetcode solution 257: Binary Tree Paths Problem Statement Given a binary tree, return all root-to-leaf paths. Note: A leaf is a node with no children. Example: Input: 1 / \ 2 3 \ 5Output: ["1->2->5", "1->3"]Explanation: All root-to-leaf paths are: 1->2->5, 1->3 P...
classSolution {publicList<String>binaryTreePaths(TreeNode root) { List<String> results =newArrayList<>();if(root ==null) {returnresults; } binaryTreePath(results,"", root);returnresults; }publicvoidbinaryTreePath(List<String>results, String path, TreeNode cur) {//直接操作字符串if(cur.left ...
Python 递归实现 # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def binaryTreePaths(self, root: TreeNode) -> List[str]: if root is None: return [] res = [] def dfs(...
*/classSolution{public://创建空容器 对象类型为string类vector<string>result;voidgetPaths(TreeNode*node,string path){if(node->left==NULL&&node->right==NULL){//左右子树为空 路径寻找完成 增加至数组中result.push_back(path);}if(node->left!=NULL){//递归遍历左子树 当前路径添加左孩子结点getPaths...
* TreeNode(int x) { val = x; } * }*/publicclassSolution {publicList<String>binaryTreePaths(TreeNode root) { List<String> res =newArrayList<String>();if(root ==null){returnres; } dfs(root,newStringBuilder(), res);returnres;
[LeetCode] 257. Binary Tree Paths Given a binary tree, return all root-to-leaf paths. Note: A leaf is a node with no children. Example: Input: 1 / \ 2 3 \ 5 Output: ["1->2->5", "1->3"] Explanation: All root-to-leaf paths are: 1->2->5, 1->3...
Given a binary tree, return all root-to-leaf paths. 递归法 复杂度 时间O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2 思路 简单的二叉树遍历,遍历的过程中记录之前的路径,一旦遍历到叶子节点便将该路径加入结果中。 代码 public class Solution { ...
treePathsIn(root.Right,strconv.Itoa(root.Val)) } return s.res } type Solution struct { res []string } func (t *Solution) treePathsIn(node *TreeNode,rootPath string) { if node.Left==nil && node.Right==nil { t.res=append(t.res,rootPath+"->"+strconv.Itoa(node.Val)) return }...
* TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicTreeNodeinvertTree(TreeNode root){if(root==null){// 节点为空,不处理returnnull;}else{// 节点不为空TreeNode leftNode;// 左节点新对象TreeNode rightNode;// 右节点新对象if(root.left!=null){leftNode=...
Tree - 257. Binary Tree Paths Given a binary tree, return all root-to-leaf paths. Note:A leaf is a node with no children. Example: 代码语言:javascript 代码 Input:1/\23\5Output:["1->2->5","1->3"]Explanation:All root-to-leaf paths are:1->2->5,1->3...