问题 给出一个二叉树,将其原地平面化为链表. 例如,给出: 1   /  \  2    5 / \     \ 3  4     6 平面化后的树看起来应该是这样: 1 \  2    \     3       \        4          \           5             \              6 初始思路 观察例子中平面化的过程,不难发现其实就是一个二叉树前序遍历的过程.让我们复习一下二叉树前序遍历的方法,根据wiki条目Tree traversal,…

1.  Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7…

目录 题目描述: 示例: 解法: 题目描述: 给定一个二叉树,原地将它展开为链表. 示例: 给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解法: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(N…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 给定一个二叉树,原地将它展开为链表. 例如,给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4…

题目 给定一个二叉树,原地将它展开为链表. 例如,给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解析 通过递归实现:可以用先序遍历,然后串成链表 主要思想就是:先递归对右子树进行链表化并记录,然后将root->right指向 左子树进行链表化后的头结点,然后一直向右遍历子树,连接上之前的右子树 /** * Definition for a binary tree node. * struct TreeNode { * int v…

114. Flatten Binary Tree to Linked List (Medium) 453. Flatten Binary Tree to Linked List (Easy) 解法1: 用stack. class Solution { public: void flatten(TreeNode *root) { if(!root) return; TreeNode *pnode = NULL; stack<TreeNode*> s; s.push(root); while(!s…

Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in th…

114 Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. 将二叉树展开成链表 [] (D:\dataStructure\Leetcode\114.png) 思路:将根节点与左子树相连,再与右子树相连.递归地在每个节点的左右孩子节点上,分别进行这样的操作. 代码 class Solution(object): def flatten(self, root): i…

Flatten a binary tree to a fake "linked list" in pre-order traversal. Here we use the right pointer in TreeNode as the next pointer in ListNode. Notice Don't forget to mark the left child of each node to null. Or you will get Time Limit Exceeded…

随笔一记,留做重温! Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 第一个想法是先序遍历,然后按照访问顺序,添加右结点. public static void…