-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathBinary_Tree_Zigzag_Level_Order_Traversal.cpp
More file actions
70 lines (64 loc) · 1.73 KB
/
Binary_Tree_Zigzag_Level_Order_Traversal.cpp
File metadata and controls
70 lines (64 loc) · 1.73 KB
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
// Source : https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal/description/
// Number : 103
// Author : HL
// Date : 2018-09-29
// Kill : 98.70%
/**********************************************************************************
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(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> result;
vector<int> v;
queue<TreeNode*> q;
TreeNode *p;
int i, s, level = 0;
if (root == NULL)
return result;
q.push(root);
while (!q.empty())
{
s = q.size();
for (i = 0; i < s; i++)
{
p = q.front();
q.pop();
v.push_back(p->val);
if (p->left != NULL)
q.push(p->left);
if (p->right != NULL)
q.push(p->right);
}
if (level % 2 != 0)
reverse(v.begin(), v.end());
level++;
result.push_back(v);
v.clear();
}
return result;
}
};