Given an n-ary tree, return the preorder traversal of its nodes' values. For example, given a 3-ary tree: Return its preorder traversal as: [1,3,5,6,2,4]. Note: Recursive solution is trivial, could you do it iteratively? 这道题让我们求N叉树的前序遍历,有之前那道Binary T…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都非常的简…
翻译 给定一个二叉树.返回其兴许遍历的节点的值. 比如: 给定二叉树为 {1. #, 2, 3} 1 \ 2 / 3 返回 [3, 2, 1] 备注:用递归是微不足道的,你能够用迭代来完毕它吗? 原文 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: R…
144. Binary Tree Preorder Traversal Difficulty: Medium Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it it…
Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? Solution: 递归…
题目:Binary Tree Preorder Traversal 二叉树的前序遍历,同样使用栈来解,代码如下: struct TreeNode { int val; TreeNode* left; TreeNode* right; TreeNode(int x): val(x), left(NULL),right(NULL) {} }; vector<int> preorderTraversal(TreeNode *root) //非递归的前序遍历(用栈实现) { if (NULL == r…
Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? 经典题目,求二叉树的后序遍历的非递归方法,跟前序,中序,层序一样都需要用到栈,后序的…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 一般我们提到树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起…
题目要求 Given an n-ary tree, return the preorder traversal of its nodes' values. 题目分析及思路 题目给出一棵N叉树,要求返回结点值的前序遍历.可以使用递归的方法做.因为是前序遍历,所以最开始就加入根结点的值. python代码 """ # Definition for a Node. class Node: def __init__(self, val, children): self.val = v…
前言   [LeetCode 题解]系列传送门:  http://www.cnblogs.com/double-win/category/573499.html   1.题目描述 Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive sol…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 二叉树的先序遍历.递归算法太简单了,重点讨论一下非递归的. 我自己写的时候一直纠结于何时弹出结点.后来看了几个版本的,发现可以跳过这个部分. 贴上我最喜欢的版本,逻辑最清楚 //我最喜欢的版本 逻辑清晰 vector<i…
Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 二叉树的前序遍历,根节点→左子树→右子树 解题思路一: 递归实现,JAVA实现如下: public List<Integer> preorderTraversal(TreeNode root) { List<I…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 求前序遍历,要求不用递归.     使用双向队列.   /** * Definition…
题目: Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 说明: 1)递归和非递归实现,其中非递归有两种方法 2)复杂度,时间O(n),空…
题目描述: Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 解题思路: 这个问题最简单的方法是使用递归,但是题目规定不能使用,得使用迭代的方法. 那么我们考虑使用栈来实现. 思路是每次遍历这节点,把该点的值放入list中,然后把该点的右孩子放入栈中,并将当前点设置为左…
Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,2,3] Follow up: Recursive solution is trivial, could you do it iteratively? ---------------------------------------------------…
Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [,,] \ / Output: [,,] Follow up: Recursive solution is trivial, could you do it iteratively? 方法一:使用迭代(C++) vector<int> preorderTraversal(TreeNode* root) { vecto…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 思路:前序遍历,“根,左子树,右子树” #include <iostream> #include <vector> #include <stack> using namespace std; s…
这道题是LeetCode里的第144道题. 题目要求: 给定一个二叉树,返回它的 前序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,2,3] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 解题代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x)…
145. Binary Tree Postorder Traversal Total Submissions: 271797 Difficulty: Hard 提交网址: https://leetcode.com/problems/binary-tree-postorder-traversal/ Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tr…
题目大意 https://leetcode.com/problems/binary-tree-inorder-traversal/description/ 94. Binary Tree Inorder Traversal Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up:…
这道题是LeetCode里的第145道题. 题目要求: 给定一个二叉树,返回它的 后序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [3,2,1] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 解题代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x)…
这道题是LeetCode里的第94道题. 题目要求: 给定一个二叉树,返回它的中序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,3,2] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 解题代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) :…
DFS,两种实现方式,递归和栈. """ # Definition for a Node. class Node: def __init__(self, val, children): self.val = val self.children = children """ class Solution: def preorder(self, root: 'Node') -> List[int]: if not root: return []…
二叉树的基础操作:二叉树的先序遍历(详细请看数据结构和算法,任意本书都有介绍),即根,左子树,右子树,实现方法中还有用栈实现的,这里不介绍了 /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution…
参考例子:[8,3,1,6,4,7,10,14,13] 8,3,1 和 6,4 说明从root开始,沿着左臂向下寻找leaf 的过程中应该逐个将node.val push入ans. class Solution(object): def preorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ stack = [] ans = [] while st…
树的先序遍历.定义一个栈,先压入中间结点并访问,然后依次压入右.左结点并访问. vector<int> preorderTraversal(TreeNode *root) { vector<int> result; stack<TreeNode *>s; TreeNode *p; p = root; s.push(p); while (!s.empty()) { p = s.top(); result.push_back(p->val); if (p->ri…
递归方法 C++代码: /* // Definition for a Node. class Node { public: int val; vector<Node*> children; Node() {} Node(int _val, vector<Node*> _children) { val = _val; children = _children; } }; */ class Solution { public: vector<int> postorder(N…
先序遍历的非递归办法,还是要用到一个stack class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ret; if(!root) return ret; stack<TreeNode*> stk; stk.push(root); //ahd(root) //a(stk) //a(ret) ){ TreeNode* cur=stk.top(); stk.p…
关于二叉树的遍历请看: http://www.cnblogs.com/stAr-1/p/7058262.html /* 考察基本功的一道题,迭代实现二叉树前序遍历 */ public List<Integer> preorderTraversal(TreeNode root) { List<Integer> res = new ArrayList<>(); if (root==null) return res; Stack<TreeNode> stack =…