Leetcode 366. Find Leaves of Binary Tree

Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

Example:
Given binary tree 

          1
         / \
        2   3
       / \     
      4   5

Returns [4, 5, 3], [2], [1].

Explanation:

1. Removing the leaves [4, 5, 3] would result in this tree: 

          1
         / 
        2

2. Now removing the leaf [2] would result in this tree:

          1

3. Now removing the leaf [1] would result in the empty tree:

          []

Returns [4, 5, 3], [2], [1].

 

解法一： 一层一层的将所有的叶子拔掉，每次去掉一个叶子，就把这个叶子的val设置成'#'。

 1 class Solution(object):
 2     def findLeaves(self, root):
 3         """
 4         :type root: TreeNode
 5         :rtype: List[List[int]]
 6         """
 7         ans = []
 8         if not root:
 9             return ans
10             
11         while root.val != '#':
12             curL = []
13             self.removeCurLeaves(root, curL)
14             ans.append([x for x in curL])
15         return ans
16         
17         
18     def removeCurLeaves(self, node, res):
19         if (not node.left or node.left.val == '#') and (not node.right or node.right.val == '#'):
20             res.append(node.val)
21             node.val = '#'
22             return 
23         if node.left and node.left.val != '#':
24             self.removeCurLeaves(node.left, res)
25         if node.right and node.right.val != '#':
26             self.removeCurLeaves(node.right, res)

 

posted @ 2017-05-10 11:25  lettuan  阅读(292)  评论(0)  编辑  收藏  举报