103.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]
]

思路：层序遍历，利用队列，一次性输出一层的所有节点，如果是偶数层，就将这一层输出的数组反转。

/**
* 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>> result;
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
if(!root)
return result;
queue<TreeNode *> q;
q.push(root);
int dep=1;
while(!q.empty()){
int size=q.size();
vector<int> level;
for(int i=0;i<size;i++){
TreeNode * node=q.front();
q.pop();
level.push_back(node->val);
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
}
if(dep%2==0)
reverse(level.begin(),level.end());
dep++;
result.push_back(level);
}
return result;
}
};