107. Binary Tree Level Order Traversal II

Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values. (i.e., from left to right, level by level from leaf to root).

Example 1:

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

ex1

Example 2:

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

Example 3:

Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -1000 <= Node.val <= 1000

Jamboard

jam

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> levels = new ArrayList<>();
        level(root, levels, 0);
        return levels;
    }
    
    void level(TreeNode node, List<List<Integer>> levels, int k) {
        if (node == null) return;
        
        if (levels.size() == k) {
            levels.add(0, new ArrayList<>());
        }
        
        level(node.left, levels, k + 1);
        level(node.right, levels, k + 1);
        
        levels.get(levels.size() - k - 1).add(node.val);
    }
}