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] ]
/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int>> result;
if(!root) return result;
vector<int> tmp;
queue<TreeNode*> q1, q2;
q1.push(root);
TreeNode *cur;
int order = 1;
while(!q1.empty()){
tmp.clear();
while(!q1.empty()){
cur = q1.front();
q1.pop();
tmp.push_back(cur->val);
if(cur -> left) q2.push(cur->left);
if(cur -> right) q2.push(cur->right);
}
if(order == 1){
result.push_back(tmp);
order = 0;
}else{
reverse(tmp.begin(), tmp.end());
result.push_back(tmp);
order = 1;
}
swap(q1, q2);
}
return result;
}
};