Find Leaves of Binary Tree 找二叉树的叶节点

Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty.

Example:
Given binary tree
1
/ \
2 3
/ \
4 5
Returns [4, 5, 3], [2], [1].

Explanation:
1. Remove the leaves [4, 5, 3] from the tree

1
/
2
2. Remove the leaf [2] from the tree

　　1

3. Remove the leaf [1] from the tree

[]
Returns [4, 5, 3], [2], [1].

该题的重点是对tree进行剪枝活动时候，我们可以利用x = change(x) 在recursive函数中。例如， root.left = removeLeaves(root.left, result); root.right = removeLeaves(root.right, result), 这样下层内嵌函数返回的null值可以赋给上层调用函数的root.left/right。 因此递归函数的返回类型很自然应该设为TreeNode

 1 public class Solution {
 2     private TreeNode removeLeaves(TreeNode root, List<Integer> result)
 3     {
 4         if(root==null)  return null;
 5         if((root.left == null)&&(root.right==null)){
 6             result.add(root.val);
 7             return null;
 8         }
 9
10         root.left = removeLeaves(root.left, result);
11         root.right = removeLeaves(root.right, result);
12
13         return root;
14
15     }
16
17     public List<List<Integer>> findLeaves(TreeNode root) {
18         List<List<Integer>> res = new ArrayList<List<Integer>>();
19         if (root==null) return res;
20         while(root!=null){
21             List<Integer> leaves = new ArrayList<Integer>();
22             root = removeLeaves(root, leaves);
23             res.add(leaves);
24         }
25
26         return res;
27     }
28 }

1
3. Remove the leaf [1] from the tree

[]
Returns [4, 5, 3], [2], [1].