标签：leetcode 解法 node append find val res 设置 blog
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)
Leetcode 366. Find Leaves of Binary Tree
标签：leetcode 解法 node append find val res 设置 blog
原文地址：http://www.cnblogs.com/lettuan/p/6228553.html
