[LeetCode] 107. Binary Tree Level Order Traversal II

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。例子，

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]
]

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

 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 };

 

DFS

 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 };