LeetCode 107. Binary Tree Level Order Traversal II

107. Binary Tree Level Order Traversal II

Difficulty: Easy

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

For example:
Given binary tree [3,9,20,null,null,15,7],

  3
 / \
9  20
  /  \
 15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

Solution

Language: Java

/**
 * 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>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode tmp = q.poll();
                level.add(tmp.val);
                if (tmp.left != null) {
                    q.offer(tmp.left);
                }
                if (tmp.right != null) {
                    q.offer(tmp.right);
                }
            }
            result.add(level);
        }
        Collections.reverse(result);
        return result;
    }
}