366. Find Leaves of Binary Tree

https://leetcode.com/problems/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 

          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].

 

 

 

Sol :

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

// The essential of problem is not to find the leaves, but group leaves of same level together and also to cut the tree. This is the exact role backtracking plays. The helper function returns the level which is the distance from its furthest subtree leaf to root, which helps to identify which group the root belongs to


public class Solution {
    public List<List<Integer>> findLeaves(TreeNode root) {
        
        List<List<Integer>> list = new ArrayList<>();
        findLeavesHelper(list, root);
        return list;
        
    }
    // return the level of root
    private int findLeavesHelper(List<List<Integer>> list, TreeNode root){
        if (root == null){
            return -1;
        }
        
        int leftLevel = findLeavesHelper(list, root.left);
        int rightLevel = findLeavesHelper(list, root.right);
        int level = Math.max(leftLevel, rightLevel) + 1;
        if (list.size() == level){
            list.add(new ArrayList<>());
        }
        list.get(level).add(root.val);
        root.left = root.right = null;
        return level;
    }
    
}