要求:Given a binary tree, flatten it to a linked list in-place.将二叉树转化为平坦序列的树.比如: 结题思路: 该题有个提示,转化后的树的序列正好是二叉树前序遍历所得到的序列,所以,该题第一个思路就是利用前序遍历的方式来做. 第二个思路:我们可以利用递归的思路,先对根节点进行处理,将root的左子树放到右子树,在将左子树中的最右端节点“嫁接”到右子树,接着上面得到的子树.接下来在用同样的方法处理当前二叉树的下一个节点.看下面一个图,即可明…

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 SOLUTION 1:使用递归解决,根据left是否为空,先连接left tree, 然后再连接右子树.使用一个t…

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 这题的主要难点是"in-place".因为这个flatten的顺序是中.左.右(相当于中序遍历),所…

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 the flattened tree, each node's right…

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 Hints: If you notice carefully in the flattened tree, each node's right child points to the…

原题地址:http://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/ 题意: 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 \ 6Hints: If you no…

题目: 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 Hints: If you notice carefully in the flattened tree, each node's right child points to…

https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/ 二叉树的处理,将二叉树转换成类似一个单链表的东东,并且原地. 类似于先跟遍历的顺序. 定义函数 TreeNode* subflatten(TreeNode *root)并将返回值设置为,先转换出来链表的最后一个位置. struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int…

题目:将一棵二叉树履平成一个类似Linked-list的东西. 思路:该过程类似于二叉树的前序遍历,但是遍历代码,我处理不来参数的变化.没AC. -------->写的很好的解题博客 参考上述博客,思路很清楚的写出代码: public void flatten(TreeNode root) { if(root == null) return; if(root.left != null){ TreeNode leftNode = root.left; TreeNode rightNode = ro…

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…