Find Leaves of Binary Tree

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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].

很有意思的一道题，主要是一步一步去除叶子结点，然后返回结果。其实这题考的是二叉树结点的高度。高度最低的先输出，同等高度的一块输出。略微修改下递归球二叉树高度的方法就可以，同时可以根据高度，确定加结果的地方。空间复杂度O(logn),时间复杂度位O(n),每个节点都需要处理一次。代码如下：

class Solution(object):
    def findLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        res = []
        self.dfs(root, res)
        return res
    def dfs(self, root, res):
        if not root:
            return -1
        left = self.dfs(root.left, res)
        right = self.dfs(root.right, res)
        level = max(left, right) + 1
        if len(res) == level:
            res.append([root.val])
        else:
            res[level].append(root.val)
        root.left = root.right = None
        return level

Find Leaves of Binary Tree

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原文地址：http://www.cnblogs.com/sherylwang/p/5667583.html