binaryTreePath(results,newArrayList<Integer>(), root);returnresults; }publicvoidbinaryTreePath(List<String> results, List<Integer>path, TreeNode cur) {if(cur ==null) {return;//root==null}if(cur.left ==null&& cur.right ==null) {//需要在上层判断 path.add(cur.val);//add cur's valStr...
排名第一的top solution是从上往下的原则,每一次把当前的path传给递归的function,这样到叶子就截止了,funcition的返回值是void的。以后遇到问题可以从上往下,从下往上都想想,参数的个数也可以多加几个,不要局限于一种思路出不来。 从上往下: publicclassSolution {publicList<String>binaryTreePaths(TreeNode root)...
self.findPaths(node.left, path, result) if node.right is not None: self.findPaths(node.right, path, result) if node.left is None and node.right is None: result.append("->".join(path)) path.pop() Node to Node Binary Tree Path 给定一棵二叉树的根节点和两个任意节点,返回这两个节点...
Go 语言递归实现 /*** Definition for a binary tree node.* type TreeNode struct {* Val int* Left *TreeNode* Right *TreeNode* }*/funcbinaryTreePaths(root*TreeNode)[]string{varres[]stringifroot==nil{returnres}dfs(root,"",&res)returnres}funcdfs(root*TreeNode,pathstring,res*[]string){...
13 void DFS(string path,TreeNode* t) 14 { 15 if(t->left==NULL&&t->right==NULL) 16 { 17 ans.push_back(path); 18 return ; 19 } 20 21 if(t->left) 22 DFS(path+"->"+to_string(t->left->val),t->left); 23 if(t->right) ...
Leetcode 144 binary tree preorder traversal 下面这种写法使用了一个辅助结点p,这种写法其实可以看作是一个模版,对应的还有中序和后序的模版写法,形式很统一,方便于记忆。 辅助结点p初始化为根结点,while循环的条件是栈不为空或者辅助结点p不为空,在循环中首先判断如果辅助结点p存在,那么先将p加入栈中,然后将...
Can you solve this real interview question? Lowest Common Ancestor of a Binary Tree - Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia [https://en.wikipedia.org/wi
LeetCode 536. Construct Binary Tree from String You need to construct a binary tree from a string consisting of parenthesis and integers. The whole input represents a binary tree. It contains an integer followed by zero, one or two pairs of parenthesis. The integer represents the root's value...
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105. 从前序与中序遍历序列构造二叉树 - 给定两个整数数组 preorder 和 inorder ,其中 preorder 是二叉树的先序遍历, inorder 是同一棵树的中序遍历,请构造二叉树并返回其根节点。 示例 1: [https://assets.leetcode.com/uploads/2021/02/19/tree.jpg] 输入: preo