【LeetCode with Python】 Flatten Binary Tree to Linked List

博客域名：http://www.xnerv.wang
原题页面：https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/
题目类型：二叉树遍历，递归回溯
难度评价：★
本文地址：http://blog.csdn.net/nerv3x3/article/details/39453299

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 child points to the next node of a pre-order traversal.

题目的算法本身并不难理解，按照先序遍历的顺序将一棵二叉树拉平为一个“链表”。因此后序遍历二叉树，先将左子树“拉平”，再将右子树“拉平”，然后如果左子树不是None，就将已经拉平为“链表”的左子树，嵌入到当前节点与直接右孩子节点（可能右孩子是None）之间。关键在于一些分支上要判断是否不为None，很容易出错。

class Solution:    # @param root, a tree node    # @return nothing, do it in place    def flatten(self, root):        if None == root:            return        if None != root.left:            self.flatten(root.left)        if None != root.right:            self.flatten(root.right)        left = root.left        right = root.left        while None != right and None != right.right:            right = right.right        if None != right:            right.right = root.right        if None != left:     ###            root.right = left        root.left = None