-
Notifications
You must be signed in to change notification settings - Fork 2
/
Copy path103. Binary Tree Zigzag Level Order Traversal.cpp
68 lines (60 loc) · 1.54 KB
/
103. Binary Tree Zigzag Level Order Traversal.cpp
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
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> res;
if(root== NULL)
return res;
queue<TreeNode*> q;
q.push(root);
int f = false;
while(q.size())
{
int size = q.size();
vector<int> level;
while(size--)
{
TreeNode* curr = q.front();
level.push_back(curr->val);
q.pop();
if(curr->left)
{
q.push(curr->left);
}
if(curr->right)
{
q.push(curr->right);
}
}
if(f)
reverse(level.begin(),level.end());
f=!f;
res.push_back(level);
}
return res;
}
};