简单思路:层序遍历
DFS 解法
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
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function (root) {
if (!root) return [];
let result = [];
dfs(root, 0, result);
return result;
};
function dfs(root, level, result) {
if (root) {
if (!result[level]) result[level] = [];
result[level].push(root.val);
dfs(root.left, level + 1, result);
dfs(root.right, level + 1, result);
}
}
BFS 算法 (队列)
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
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function (root) {
if (!root) return [];
let queue = [root];
let result = [];
while (queue.length > 0) {
const res = [];
let len = queue.length;
while (len) {
const node = queue.shift();
res.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
len--;
}
result.push(res);
}
return result;
};