144. Binary Tree Preorder Traversal 前序的非递归遍历:用堆来实现 如果把这个代码改成先向堆存储左节点再存储右节点,就变成了每一行从右向左打印 如果用队列替代堆,并且先存储左节点,再存储右节点,就变成了逐行打印 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> result; if(root == NULL) return res…
144. 二叉树的前序遍历 144. Binary Tree Preorder Traversal 题目描述 给定一个二叉树,返回它的 前序 遍历. LeetCode144. Binary Tree Preorder Traversal 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? Java 实现 Iterative Solution import java.util.LinkedList; import…
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…
原题 Binary Tree Preorder Traversal 没什么好说的... 二叉树的前序遍历,当然如果我一样忘记了什么是前序遍历的..  啊啊.. 总之,前序.中序.后序,是按照根的位置来决定的,根首先访问,是前序. /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(N…
二叉树的非递归前序遍历,大抵是很多人信手拈来.不屑一顾的题目罢.然而因为本人记性不好.基础太差的缘故,做这道题的时候居然自己琢磨出了一种解法,虽然谈不上创新,但简单一搜也未发现雷同,权且记录,希望于人于己皆有帮助. 估计能看到这里的读者,都是见过题目的,但还是链接原题在此:Binary Tree Preorder Traversal 二话不说,直接上代码: /** * Definition for binary tree * public class TreeNode { * int val;…
详见:剑指 Offer 题目汇总索引:第6题 Binary Tree Postorder Traversal            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…
Binary Tree Preorder Traversal My Submissions QuestionEditorial Solution Total Accepted: 119655 Total Submissions: 300079 Difficulty: Medium Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,…
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]. c++版: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNod…
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? 解法一:递归 /** *…
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    /   3return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively?…
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…
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…
先序遍历,用递归来做,简单的不能再简单了.代码如下: (以下仅实现了先序遍历,中序遍历类似,后序遍历和这两个思路不一样,具体详见Binary Tree Postorder Traversal) /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), righ…
Total Accepted: 97599 Total Submissions: 257736 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…
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? 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…
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]. 前序遍历二叉树,只不过题目的要求是尽量不要使用递归,当然我还是先用递归写了一个: /** * Definition for a binary tree node. * struct TreeNode { * int val…
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 给定一个二叉树,返回他的前序遍历的节点的values. 例如: 给定一个二叉树 {1,#,2,3}, 1 \ 2 / 3 返回 [1,2,3]. 笔记: 递归解决方案是微不足道的,你可以用迭代的方法吗? +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++…
题目: 二叉树的前序遍历 给出一棵二叉树,返回其节点值的前序遍历. 样例 给出一棵二叉树 {1,#,2,3}, 1 \ 2 / 3 返回 [1,2,3]. 挑战 你能使用非递归实现么? 解题: 通过递归实现,根节点->左节点->右节点 Java程序: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(…
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,null,2,3], 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 递归 : class Solution { private List<Intege…
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)…
给定一棵二叉树,返回其节点值的前序遍历.例如:给定二叉树[1,null,2,3],   1    \     2    /   3返回 [1,2,3].注意: 递归方法很简单,你可以使用迭代方法来解决吗?详见:https://leetcode.com/problems/binary-tree-preorder-traversal/description/ Java实现: 递归实现: /** * Definition for a binary tree node. * public class T…
题目描述 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? 解题思路 由于前序遍历的顺序是父节点->左孩子->右孩子,所以在遍历到某节点时,先输出该节点值,然后把该节点的右孩子入栈,接着访问左孩子,若左孩子为空,则访问栈顶结点. 代码 /** * Definition for a binary tree node. * struct TreeNode { * int…
[题目] Given a binary tree, return the preordertraversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,2,3] [思路] 有参考,好机智,使用堆栈压入右子树,暂时存储. 左子树遍历完成后遍历右子树. [代码] /** * Definition for a binary tree node. * public class TreeNode { *…
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? 递归: class Solution { public: vector<int> res; vector<int> preorderTraversal(TreeNode* root) { GetAns(root); return res; } void GetAns(TreeNode* root) { if…
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 =…
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…
题意:给一棵树,求其先根遍历的结果. 思路: (1)深搜法: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> preorderT…
题目意思:二叉树先序遍历,结果存在vector<int>中 解题思路:1.递归(题目中说用递归做没什么意义,我也就贴贴代码吧) 2.迭代 迭代实现: class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ans; if(root){ TreeNode* temp; stack<TreeNode*> s; //利用栈,每次打印栈顶,然后将栈顶弹出…