原题:
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
思路:
由上面可以看出:这道题的意思是将一颗二叉树平化(flatten)为一条链表,而链表的顺序为二叉树的先序遍历。
解题思路:首先将左右子树分别平化为链表,这两条链表的顺序分别为左子树的先序遍历和右子树的先序遍历。然后将左子树链表插入到根节点和右子树链表之间,就可以了。左右子树的平化则使用递归实现。
代码:
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# @param root, a tree node
# @return nothing, do it in place
def flatten(self, root):
if root == None:
return
self.flatten(root.left)
self.flatten(root.right)
p = root
if p.left == None:
return
p = p.left
while p.right:
p = p.right
p.right = root.right
root.right = root.left
root.left = None