[LeetCode] 107. Binary Tree Level Order Traversal II

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]
]

二叉树层序遍历二。题意跟版本一很接近，唯一的不同点是输出的output需要从下往上排列，叶子节点层最先输出，根节点在最后输出。

因为跟版本一很接近我就不过多解释思路了，依然是BFS和DFS两种做法。两种做法的时间和空间复杂度均为O(n)。

BFS

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {number[][]}
 4  */
 5 var levelOrderBottom = function (root) {
 6     let res = [];
 7     if (root === null) return res;
 8     let queue = [];
 9     queue.push(root);
10     while (queue.length) {
11         let size = queue.length;
12         let list = [];
13         for (let i = 0; i < size; i++) {
14             let cur = queue.shift();
15             if (cur.left !== null) queue.push(cur.left);
16             if (cur.right !== null) queue.push(cur.right);
17             list.push(cur.val);
18         }
19         res.unshift(list);
20     }
21     return res;
22 };

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public List<List<Integer>> levelOrderBottom(TreeNode root) {
12         List<List<Integer>> res = new ArrayList<>();
13         // corner case
14         if (root == null) {
15             return res;
16         }
17 
18         // normal case
19         Queue<TreeNode> queue = new LinkedList<>();
20         queue.offer(root);
21         while (!queue.isEmpty()) {
22             int size = queue.size();
23             List<Integer> list = new ArrayList<>();
24             for (int i = 0; i < size; i++) {
25                 TreeNode cur = queue.poll();
26                 list.add(cur.val);
27                 if (cur.left != null) {
28                     queue.offer(cur.left);
29                 }
30                 if (cur.right != null) {
31                     queue.offer(cur.right);
32                 }
33             }
34             res.add(0, list);
35         }
36         return res;
37     }
38 }

DFS

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {number[][]}
 4  */
 5 var levelOrderBottom = function (root) {
 6     let res = [];
 7     if (root === null) return res;
 8     helper(res, root, 0);
 9     return res;
10 };
11 
12 var helper = function (res, root, level) {
13     if (root === null) return;
14     if (level >= res.length) {
15         res.unshift([]);
16     }
17     res[res.length - level - 1].push(root.val);
18     helper(res, root.left, level + 1);
19     helper(res, root.right, level + 1);
20 };