postorder专题

PreOrder, InOrder, PostOrder 题型总结

最基本的 Preorder, Inorder, Postorder Traverse Binary Tree Preorder traverse (用stack模拟，先进去right ，再进去left） Binary Tree Inorder traverse （用stack模拟，每次push整个左枝的所有node，再变换到右边); Binary Tree Postorder travers

LeetCode | Binary Tree Postorder Traversal

后序遍历，当然，，，用栈实现后序遍历的关键在于从左孩子回到根节点和从右孩子回到跟节点。如果是从左回的，要继续向右。如果是从右边回来的（表明左右孩子已经遍历结束）这时候就需要将自身输出。具体到代码上，就是使用一个q指针，用于保存上一个走过的节点（确切的说应该是输出的节点）。然后从栈顶取出元素的时候，判断p->right==q? 来决定它是从哪个方向回来的~ 完成使命~~ 注释里面

Binary Tree Postorder Traversal-二叉树的后序遍历

原题： Given a binary tree, return the postorder traversal of its nodes' values. =>给定一个二叉树，给出后序遍历的所有节点值 For example: =>例如 Given binary tree {1,#,2,3}, =>给定一个二叉树{1，#，2,3} 1\2/3 return [3,2,1

leetcode No106. Construct Binary Tree from Inorder and Postorder Traversal

Question： Given inorder and postorder traversal of a tree, construct the binary tree. 根据树的中序遍历和后序遍历构建二叉树 Algorithm：中序遍历：左-根-右后序遍历：左-右-根举个例子中序遍历：DBEAFCG 后序遍历：DEBFGCA 1、后序遍历的最后一

[LeetCode] Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes’ values. For example: Given binary tree {1,#,2,3}, return [3,2,1]. Note: Recursive solution is trivial, could you do it iterative

Leetcode 106 Construct Binary Tree from Inorder and Postorder Traversal（二叉树）

题目连接：Leetcode 106 Construct Binary Tree from Inorder and Postorder Traversal 解题思路：参考 Leetcode 105。 /*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;*

LeetCode 题解（155）: Construct Binary Tree from Inorder and Postorder Traversal

题目： Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 题解：递归。 C++版： class Solution {public:TreeNode*

***Leetcode 145. Binary Tree Postorder Traversal

https://leetcode.com/problems/binary-tree-postorder-traversal/description/ 给一种容易理解的写法： class Solution {public:vector<int> postorderTraversal(TreeNode* root) {stack<TreeNode*> sta;unordered_set< Tre

68.Binary Tree Postorder Traversal-二叉树的后序遍历（容易题）

二叉树的后序遍历题目给出一棵二叉树，返回其节点值的后序遍历。样例给出一棵二叉树 {1,#,2,3}, 1 \ 2 / 3 返回 [3,2,1]挑战你能使用非递归实现么？题解 1.递归法 /*** Definition of TreeNode:* public class TreeNode {* public int val;* public TreeNo

LeetCode - construct-binary-tree-from-inorder-and-postorder-traversal

题目： Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 题意：给你一个二叉树的中序序列和后序序列，确定二叉树。里面的结点不重复解题思路：

LeetCode 145 Binary Tree Postorder Traversal (后序遍历二叉树)

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

LeetCode 106：Construct Binary Tree from Postorder and Inorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree. 给定一个二叉树的后序和中序遍历，重建这棵二叉树。此题和LeetCode105 根据前序和中序重建二叉树类似。所谓后序遍历，即先访问根的左、右子树，然后再访问根节点。这样根节点在二叉树后序遍历的最后一个个元素。所谓中序遍历

leetcode106~Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据中序遍历和后序遍历创建一棵二叉树。思路与上题类似。关键在于，上下限的选取。下面两种方法，不同在于上下限选

106. Construct Binary Tree from Inorder and Postorder Traversal【M】【31】

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. Subscribe to see which companies asked this

590. N-ary Tree Postorder Traversal

590. N叉树的后序遍历给定一个 N 叉树，返回其节点值的后序遍历。例如，给定一个 3叉树 : 返回其后序遍历: [5,6,3,2,4,1]. 说明: 递归法很简单，你可以使用迭代法完成此题吗? 解法一 /*// Definition for a Node.class Node {public:int val;vector<Node*> children

leetcode--Binary Tree Postorder Traversal

题目描述： 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

Leetcode#145. Binary Tree Postorder Traversal

题目描述：二叉树的后序遍历，打印出结点值，返回一个数组解题思路：递归实现如下： C++版本： class Solution {public:vector<int> helper(TreeNode* root, vector<int> &r){if(root){helper(root->left, r);helper(root->right, r);r.push_back(root->va

16 二叉树的后序遍历（Binary Tree Postorder Traversal）

文章目录 1 题目2 描述3 解决方案3.1 递归算法3.1.1 遍历法（Traverse）思路源码 3.1.2 分治法（Devide And Conquer）思路源码 3.2 非递归算法3.2.1 二叉树遍历的非递归通用解法思路源码图解 3.3 时间复杂度3.4 空间复杂度 1 题目二叉树的后序遍历（Binary Tree Postorder Traversal）

Leetcode 145. Binary Tree Postorder Traversal二叉树后序遍历

Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [1,null,2,3]1\2/3Output: [3,2,1] 题目链接:https://leetcode.com/problems/binary-tree-postorder-traversal/ /***

LeetCode_Construct Binary Tree from Inorder and Postorder Traversal

给定二叉树中序、后序遍历结果，构建二叉树。后序的最后一个结点必为根节点。对于后序的最后一个结点而言，中序遍历结果中，位于它左边的是它的左子树，位于它右边的是它的右子树。由 2 可得基本递归思路一：不断利用后序最后一个节点将中序结果分割，自底向上递归建立左右子树即可。此时，由于要在中序结果中寻找后序最后结点，故要么每次都循环遍历（时间代价），要么提前建立一个哈希表（空间代价）作为参数。本文的重点