[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]
]

/**
* 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:
/*algorithm: BFS
use stack to reverse instead of reverse
*/
void bfs(vector<vector<int>>&result,TreeNode* root){
queue<TreeNode*>Q;
if(root)Q.push(root);
bool flip = false;
while(!Q.empty()){
int size = Q.size();
vector<int>path;
stack<int>stk;
for(int i = 0;i < size;i++){
TreeNode* t = Q.front();Q.pop();
if(t->left)Q.push(t->left);
if(t->right)Q.push(t->right);
if(flip)stk.push(t->val);
else path.push_back(t->val);
}
if(flip){
while(!stk.empty()){
path.push_back(stk.top());
stk.pop();
}
}
flip = !flip;
result.push_back(path);
}
}
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>>result;
vector<int>path;
bfs(result,root);
return result;
}
};

/**
* 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:
/*algorithm:bfs recursive
*/
void preorder(vector<vector<int>>&result,TreeNode* root,int level){
if(!root)return;
if(result.size() <= level)
result.push_back(vector<int>());
if(level%2)//odd
result[level].insert(result[level].begin(),root->val);
else
result[level].push_back(root->val);
preorder(result,root->left,level+1);
preorder(result,root->right,level+1);
}
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>>result;
preorder(result,root,0);
return result;
}
};