[LeetCode] 366. 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].

Thought process:
The question asks us to "collect a tree's nodes as if you were doing this: collect and remove all leaves, repeat until the tree is empty". So we don't actually remove the leaves. Instead, we add a node to its correct list by calculating its height.

Solution:

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/**
 * 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>> leaves = new ArrayList<>();
        
        findLeavesByHeight(root, leaves);
        
        return leaves;
    }
    
    private int findLeavesByHeight(TreeNode root, List<List<Integer>> leaves) {
        if (root == null) {
            return -1;
        }
        
        int height = Math.max(findLeavesByHeight(root.left, leaves), findLeavesByHeight(root.right, leaves)) + 1;
        
        if (leaves.size() <= height) {
            leaves.add(new ArrayList<>());
        }
        leaves.get(height).add(root.val);
        
        return height;
    }
}

Time complexity:
Say n = number of nodes, the time complexity is O(n).
Location: San Jose, CA, USA

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