Problem

Given a binary tree, return the bottom-up level order traversal of its nodes’ values.

Examples

Example 1:

    3
   / \
  9   20
      /  \
     15   7

Input: root = [3,9,20,null,null,15,7]
Output: [ [15,7],[9,20],[3] ]

Example 2:

Input: root = [1]
Output: [ [1] ]

Example 3:

Input: root = []
Output: []

This is follow up of Binary Tree Level Order Traversal - Level by Level

Solution

Method 1 - Use BFS and reverse

We have Idea is to take the level and add to new list.

public List <ArrayList <Integer>> levelOrderBottom(TreeNode root) {
    ArrayList <ArrayList<Integer>> result = new ArrayList<ArrayList <>>();

    if (root == null) {
        return result;
    }

    LinkedList <TreeNode> current = new LinkedList <TreeNode> ();
    LinkedList <TreeNode> next = new LinkedList <TreeNode> ();
    current.offer(root);

    ArrayList <Integer> numberList = new ArrayList <Integer> ();

    // need to track when each level starts
    while (!current.isEmpty()) {
        TreeNode head = current.poll();

        numberList.add(head.val);

        if (head.left != null) {
            next.offer(head.left);
        }
        if (head.right != null) {
            next.offer(head.right);
        }

        if (current.isEmpty()) {
            current = next;
            next = new LinkedList <TreeNode> ();
            result.add(numberList);
            numberList = new ArrayList <Integer> ();
        }
    }

    //return Collections.reverse(result);
    List <List <Integer>> reversedResult = new ArrayList <ArrayList <Integer>> ();
    for (int i = result.size() - 1; i>= 0; i--) {
        reversedResult.add(result.get(i));
    }

    return reversedResult;
}