LeetCode 366. 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:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> findLeaves(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        height(root, res);
        return res;
    }
    
    private int height(TreeNode root, List<List<Integer>> res){
        if(root == null){
            return -1;
        }
        int h = 1+Math.max(height(root.left, res), height(root.right, res));
        if(res.size() < h+1){
            res.add(new ArrayList<Integer>());
        }
        res.get(h).add(root.val);
        return h;
    }
}

类似Maximum Depth of Binary Tree. 

跟上Boundary of Binary Tree.