LeetCode Find Leaves of Binary Tree

原题链接在这里：https://leetcode.com/problems/find-leaves-of-binary-tree/#/description

题目：

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].

题解：

计算当前node到leaf的距离为当前node的高度，相同高度的点放在同一个list里.

Time Complexity: O(n). leaf层用时n/2, leaf上一层用时n/4*2, 再上一层用时n/8*3. 是n(i/2^i)的和. i = 1,2,3....

Space: O(logn), stack space. Regardless res.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<List<Integer>> findLeaves(TreeNode root) {
12         List<List<Integer>> res = new ArrayList<List<Integer>>();
13         height(root, res);
14         return res;
15     }
16
17     private int height(TreeNode root, List<List<Integer>> res){
18         if(root == null){
19             return -1;
20         }
21         int h = 1+Math.max(height(root.left, res), height(root.right, res));
22         if(res.size() < h+1){
23             res.add(new ArrayList<Integer>());
24         }
25         res.get(h).add(root.val);
26         return h;
27     }
28 }

类似Maximum Depth of Binary Tree.