*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：https://discuss.leetcode.com/topic/49400/1-ms-easy-understand-java-solution

O(n^2)

DFS

/**
 * 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>> leavesList = new ArrayList< List<Integer>>();
        List<Integer> leaves = new ArrayList<Integer>();
        while(root != null) {
            if(isLeave(root, leaves)) root = null;
            leavesList.add(leaves);
            leaves = new ArrayList<Integer>();
        }
        return leavesList;
    }
    
    public boolean isLeave(TreeNode node, List<Integer> leaves) {
        if (node.left == null && node.right == null) {
            leaves.add(node.val);
            return true;
        }
        if (node.left != null) {
             if(isLeave(node.left, leaves))  node.left = null;
        }
        if (node.right != null) {
             if(isLeave(node.right, leaves)) node.right = null;
        }
        return false;
    }
}