Binary Tree Zigzag Level Order Traversal

Desicription

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],

1
2
3
4
5
3
/ \
9 20
/ \
15 7

return its zigzag level order traversal as:

1
2
3
4
5
[
[3],
[20,9],
[15,7]
]

Solution

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
/**
* 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 {
private:
vector<vector<int>> res;
void buildVec(TreeNode* root, int level) {
if(root == NULL)
return ;
if(level == res.size())
res.push_back(vector<int>());
res[level].push_back(root->val);
buildVec(root->left, level+1);
buildVec(root->right, level+1);
}
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
buildVec(root, 0);
for(int i = 0; i < res.size(); i++) {
if(i & 1) {
for(int left = 0, right = res[i].size() - 1; left <= right; left++, right--) {
swap(res[i][left], res[i][right]);
}
}
}
return res;
}
};