-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathindex.js
70 lines (64 loc) · 1.5 KB
/
index.js
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
// Title : binary-tree-level-order-traversal
// Date : 2019-03-14
// Author : Daguo
/*****************************************
给定一个二叉树,返回其按层次遍历的节点值。 (即逐层地,从左到右访问所有节点)。
例如:
给定二叉树: [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其层次遍历结果:
[
[3],
[9,20],
[15,7]
]
*****************************************/
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
// 循环,队列
var levelOrder = function(root) {
if (root === null) return [];
// 存放每一层的节点
let queue = [root];
let result = [];
while (queue.length) {
let cur = [];
// 存放每一层的节点值
let values = [];
queue.forEach(v => {
if (v.val !== null) values.push(v.val);
if (v.left) cur.push(v.left);
if (v.right) cur.push(v.right);
});
result.push(values);
queue = cur;
}
return result;
};
// 递归,左树一定比右树先存储,保证了顺序
var levelOrder2 = function(root) {
let result = [];
const find = (root, depth, res) => {
if (!root) return;
if (depth >= res.length) res.push([]);
res[depth].push(root.val);
find(root.left, depth + 1, res);
find(root.right, depth + 1, res);
};
find(root, 0, result);
return result;
};
module.exports = [levelOrder, levelOrder2];