366. Find Leaves of Binary Tree

Description

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

tree

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

Explanation:

1. Removing the leaves [4, 5, 3] would result in this tree:

image.png

2. Now removing the leaf [2] would result in this tree:

image.png

3. Now removing the leaf [1] would result in the empty tree:

image.png

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

Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.

Solution

DFS, time O(n), space O(n)

利用Depth辅助即可。depth定义成从最深的子节点到自己的距离，那么所有最外层节点的depth都是0，次外层节点的depth为1，以此类推。

起先用的是HashMap + PriorityQueue的解法，后来发现并不需要，用List<List<Integer>>即可，因为一定是从外到里添加到结果集，顺序不会被打乱。

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