题目：
Given a binary tree, flatten it to a linked list in-place.
算法思路：
其实该题目就是二叉树前序遍历的变形
我代码沿用leetcode 144. Binary Tree Preorder Traversal
代码：
class Solution(object): def preorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ if root == None:return [] stack = [root] result = [] while len(stack) != 0: tmp_root = stack.pop() if tmp_root == None:continue result.append(tmp_root) stack.append(tmp_root.right) stack.append(tmp_root.left) return result def flatten(self, root): """ :type root: TreeNode :rtype: void Do not return anything, modify root in-place instead. """ result = self.preorderTraversal(root) for i in range(1,len(result)): result[i-1].left = None result[i-1].right = result[i]
leetcode 114. Flatten Binary Tree to Linked List （Python版）
原文：http://wdswds.blog.51cto.com/11139828/1737957