4.2 leetcode -2 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}".

这个题目就是让我实战一次分层遍历的，然后至于Z字型，只需要翻转就行了

/**
* 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) {
queue<TreeNode*> node;
vector<vector<int> > vreturn;

bool flag = true;//0-left-right
int ceil_num = 0;
int ceil_numtmp = 0;
if(root != NULL )
{
ceil_num = 1;
node.push(root);

}
else
return vreturn;

while(!node.empty())
{
vector<int> numvector(ceil_num);

for(int i = 0;i < ceil_num;i++)
{
TreeNode *now = node.front();
numvector[i] = now->val;
node.pop();
if(now->left != NULL)
{
ceil_numtmp ++;
node.push(now->left);
}
if(now->right != NULL)
{
ceil_numtmp ++;
node.push(now->right);
}
}
flag = !flag;
if(flag == true)
{
//reverse
for(int i = 0;i < (ceil_num/2);i++)
{
int tmp = numvector[ceil_num -i - 1];
numvector[ceil_num -i - 1] = numvector[i];
numvector[i] = tmp;
}
}
vreturn.push_back(numvector);
ceil_num = ceil_numtmp;
ceil_numtmp = 0;
}
return vreturn;
}
};