【LeetCode】Binary Tree Zigzag Level Order Traversal

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,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

[
[3],
[20,9],
[15,7]
]

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

.

思路：BFS，对偶数层进行逆序。

/**
* Definition for binary tree
* 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> > ret;
vector<int> rettmp;
queue<TreeNode *> level;
TreeNode *tmp = root;
int i = 0;
level.push(root);
if(NULL == root)return ret;
level.push(NULL);
while(!level.empty()){
tmp = level.front();
level.pop();
if(NULL == tmp){
if(0 == (++i)%2){
reverse(rettmp.begin(), rettmp.end());
}
ret.push_back(rettmp);
rettmp.clear();
if(!level.empty())level.push(NULL);
}else{
if(NULL != tmp->left){
level.push(tmp->left);
}
if(NULL != tmp->right){
level.push(tmp->right);
}
rettmp.push_back(tmp->val);
}
}
return ret;
}
};