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
要求是in-place的修改, 所以只能通过遍历和修改左右节点的引用来实现, 提示中说如果是前序遍历的话, 右边节点的前一个节点就是需要连接过去的目标.
照着这个思路, 需要在递归中获取的信息有: 当前节点是否是右节点, 当前节点的父节点. 然后通过处理变成一棵线性, 全部是左节点的树.
处理顺序是, 如果节点是右节点, 先把它添加为前一个节点的左节点, 然后修改父节点, 将父节点的右节点置为空.
最后将左倾树变化为目标形式.
python 代码:
1 class Solution(object): 2 def __init__(self): 3 self.prev = None 4 5 def flatten(self, root): 6 """ 7 :type root: TreeNode 8 :rtype: void Do not return anything, modify root in-place instead. 9 """ 10 self.deal(root, None, False) 11 self.moveLeftToRight(root) 12 13 def deal(self, node, parent, isRight): 14 if not node: 15 return 16 if isRight: 17 parent.right = None 18 self.prev.left = node 19 20 self.prev = node 21 self.deal(node.left, node, False) 22 self.deal(node.right, node, True) 23 24 def moveLeftToRight(self, node): 25 while node: 26 left = node.left 27 node.left = None 28 node.right = left 29 node = left 30
后来翻看别人的解答时, 发现了一个很棒的方法.
思路是: reversed pre-order, 从右往左遍历, 已demo为例顺序是 6->5->4->3->2->1, 记录下前一个节点后, 将其添加为当前节点的右节点, 然后左节点为空. 递归完毕后就是一棵右倾树了.
1 def __init__(self): 2 self.prev = None 3 4 def flatten(self, root): 5 if not root: 6 return None 7 self.flatten(root.right) 8 self.flatten(root.left) 9 10 root.right = self.prev 11 root.left = None 12 self.prev = root
https://discuss.leetcode.com/topic/33192/8-lines-of-python-solution-reverse-preorder-traversal/2
(完)