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? --------------------------------------------------…

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…

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]. 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…

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. 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]. 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? 例如以下列出三种解法:一种迭代的和两种iterative的 import j…

题目: 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? 分析: 给定一棵二叉树,返回后序遍历. 递归方法很简单,即先访问左子树,再访问右子树,最后访…

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)…

采用广度优先遍历,一个变量记录层数,一个变量记录方向. void traverse(TreeNode *root, vector<vector<int>> result, int level, bool left_to_right) { if (!root)return; //如果进入下一层了,则result同样也加一层 if (level > result.size())result.push_back(vector<int>()); //如果此时为从左向右,则…

二叉树的后序遍历 用标记右子树vector的方法 vector<int> postorderTraversal(TreeNode *root) { vector<int> ans; vector<TreeNode *> stack; vector<bool> isRight; stack.push_back(root); isRight.push_back(false); TreeNode * pNode = NULL; while(!stack.empty…

原题地址 递归写法谁都会,看看非递归写法. 对于二叉树的前序和中序遍历的非递归写法都很简单,只需要一个最普通的栈即可实现,唯独后续遍历有点麻烦,如果不借助额外变量没法记住究竟遍历了几个儿子.所以,最直接的想法就是在栈中记录到底遍历了几个儿子. 代码: vector<int> postorderTraversal(TreeNode *root) { vector<int> path; stack<pair<TreeNode *, int> > st; st.p…

中序遍历二叉搜索树,得到的是一个有序的结果,找出其中逆序的地方就可以了.如果逆序的地方相邻,只需把逆序的相换即可:如果不相邻,则需要找到第二个逆序对的 第二个元素再做交换. 定义两个指针p和q来指定需要交换的元素,指针pre记录当前结点的前驱结点,用来判断是否逆序. void recoverTree(TreeNode *root) { pre = p = q = nullptr; dfs(root); swap(p->val, q->val); } void dfs(TreeNode *roo…

层序遍历,使用队列将每层压入,定义两个队列来区分不同的层. vector<vector<int>> levelorderTraversal(TreeNode *root) { vector<vector<int>> result; vector<int>tmp; //通过两个queue来区分不同的层次 queue<TreeNode *>current,next; TreeNode *p = root; current.push(p);…

后序遍历,比先序和中序都要复杂.访问一个结点前,需要先判断其右孩子是否被访问过.如果是,则可以访问该结点:否则,需要先处理右子树. vector<int> postorderTraversal(TreeNode *root) { vector<int> result; stack<TreeNode *>s; TreeNode *p, *q;//一个表示当前访问的结点,一个表示刚刚访问过的结点 p = root; do { while (p != nullptr) { /…

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…

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 /**…

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…

题目:Binary Tree Postorder Traversal 二叉树的后序遍历,题目要求是采用非递归的方式,这个在上数据结构的课时已经很清楚了,二叉树的非递归遍历不管采用何种方式,都需要用到栈结构作为中转,代码很简单,见下: struct TreeNode { int val; TreeNode* left; TreeNode* right; TreeNode(int x): val(x), left(NULL),right(NULL) {} }; vector<int> preord…

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…

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…

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

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…

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…