LeetCode - Binary Tree Level Order Traversal II

題目：

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

思路：

遞歸，然後swap

package treetraversal;

import java.util.ArrayList;
import java.util.List;

public class BinaryTreeLevelOrderTraversalII {
    
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        levelOrderTraversal(res, root, 0);
        int n = res.size();
        for (int i = 0; i < n / 2; ++i) {
            List<Integer> tmp = res.get(i);
            res.set(i, res.get(n - i - 1));
            res.set(n - i - 1, tmp);
        }
        return res;
    }
    
    private void levelOrderTraversal(List<List<Integer>> res, TreeNode root, int k) {
        if (root == null) return;
        if (res.size() < k + 1) {
            List<Integer> record = new ArrayList<Integer>();
            record.add(root.val);
            res.add(record);
        } else {
            res.get(k).add(root.val);
        }
        levelOrderTraversal(res, root.left, k + 1);
        levelOrderTraversal(res, root.right, k + 1);
    }
    
}