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? 这道题想了很久,后序遍历应该是最麻烦的吧,最容易想到的方法就是直接设置标志位. 相比…

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 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…

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…

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    /   3return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively…

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 you do it iteratively? 解法一:递归法 /**…

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…

145. 二叉树的后序遍历 145. Binary Tree Postorder Traversal 题目描述 给定一个二叉树,返回它的 后序 遍历. LeetCode145. Binary Tree Postorder Traversal 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? Java 实现 Iterative Solution import java.util.LinkedList; impo…

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,后续的…

题目: 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? 链接: http://leetcode.com/problems/binary…

Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [3,2,1] Follow up: Recursive solution is trivial, could you do it iteratively? --------------------------------------------------…

题目: 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? 题解: 递归方法代码:          ArrayList<Integer…

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…

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 postorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right;…

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? vector<int> postorderTraversal(TreeNode *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]. 后序遍历,左子树→右子树→根节点 前序遍历的非递归实现需要一个计数器,方法是需要重写一个类继承TreeNode,翁慧玉教材<数据结构:题解与拓展>P113有详细介绍,这里略.递归JAVA实现如下: public L…

Problem Link: http://oj.leetcode.com/problems/binary-tree-postorder-traversal/ The post-order-traversal of a binary tree is a classic problem, the recursive way to solve it is really straightforward, the pseudo-code is as follows. RECURSIVE-POST-ORDE…

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? 求后序遍历,要求不使用递归. 使用栈,从后向前添加. /** * Definition…

题目: 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? 说明: 1) 两种实现,递归与非递归 , 其中非递归有两种方法 2)复杂度分析…

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? 1.递归算法的递归定义是: 若二叉树为空,则遍历结束:否则⑴ 后序遍历左子树(递归调用…

题目: Given a list, rotate the list to the right by k places, where k is non-negative. For example:Given 1->2->3->4->5->NULL and k = 2,return 4->5->1->2->3->NULL. Given a binary tree, return the postorder traversal of its nodes…

Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [3,2,1] Follow up: Recursive solution is trivial, could you do it iteratively? 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…

Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [,,] \ / Output: [,,] Follow up: Recursive solution is trivial, could you do it iteratively? 方法一:利用两个栈s1,s2来实现,先将头结点入栈s1,从s1弹出栈顶节点记为cur,压入s2中,分别将cur的左右孩子压入s1,当s…

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? Subscribe to see which companies asked this…

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]. 思路:后序遍历是按照“左子树,右子树,根”的顺序访问元素.那么根或者其它父亲元素就要先压入栈,然后再弹出. #include <iostream> #include <algorithm> #includ…

题目: 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? 解题思路: 后序遍历的非递归实现与前序遍历和中序遍历的非递归不同,对于根节点…

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]. 解题: 二叉树的后序遍历,比先序和中序遍历稍显麻烦一些,有兴趣可查看:二叉树先序遍历.二叉树中序遍历 后序遍历需要输出左子树后,输出右子树,最后输出当前结点.复杂的原因是,输出左右子树后,如何判断输出的结点是左还是右.…

Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tree [1,null,2,3], 1 \ 2 / 3 return [3,2,1]. 递归: class Solution { List<Integer> res = new ArrayList<Integer>(); public List<Integer> posto…

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? class Solution { public: vector<int> post…