LeetCode（26）-Binary Tree Level Order Traversal II

题目：

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：

• 递归的思路
-

代码：

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */
public class Solution {
   public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        levelOrderBottomHelper(root, result, 1);
        return result;
    }

    public void levelOrderBottomHelper(TreeNode root, List<List<Integer>> result, int depth) {
        if (root == null)
            return;
        List<Integer> list;
        if (result.size() < depth) {
            list = new ArrayList<Integer>();
            result.add(0, list);
        } else {
            list = result.get(result.size() - depth);
        }
        list.add(root.val);
        levelOrderBottomHelper(root.left, result, depth+1);
        levelOrderBottomHelper(root.right, result, depth+1);
    }
}