[LeetCode] Binary Tree Zigzag Level Order Traversal（bfs）

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) {
vector<vector<int> > res;
if(root==NULL)
return res;
queue<TreeNode *> q;
q.push(root);
q.push(NULL);
res = bfs(q);
return res;
}
private:
vector<vector<int> > bfs(queue<TreeNode *> q){
vector<vector<int> > res;
vector<int> temp;
bool needReverse = false;

while(!q.empty()){
TreeNode *p = q.front();
q.pop();
if(p!=NULL){
temp.push_back(p->val);
if(p->left!=NULL){
q.push(p->left);
}
if(p->right!=NULL){
q.push(p->right);
}
}else if(p==NULL ){

if(needReverse){
int len = temp.size();
for(int i=0,j=len-1;i<j;i++,j--)
swap(temp[i],temp[j]);
}
res.push_back(temp);
temp.clear();
needReverse = (!needReverse);
if(!q.empty())
q.push(NULL);
}

}//end while
return res;
}//end bfs
};

方法：用queue实现bfs，对树按层进行搜索，树的结点按层存入queue中，用NULL值标记本层的结束。用bool值按层取反决定本层的vector是否要逆置。