[Leetcode] Binary tree Zigzag level order traversal二叉树Z形层次遍历

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}". 题目要求按Z字形以此输出每层，主体还是层次遍历。 思路一：只在Binary tree level order traversal的基础上增加对层数的奇偶的判断，然后依此来决定是否反转当前行即可；

/**
* 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>> levelOrderBottom(TreeNode* root)
{
vector<vector<int>> res;
queue<TreeNode *> Q;
if(root)    Q.push(root);
size_t level=1;     //第level层，初始为第一层
while( !Q.empty())
{
int count=0;
int levCount=Q.size();
vector<int> levNode;

while(count<levCount)
{
TreeNode *curNode=Q.front();
Q.pop();
levNode.push_back(curNode->val);
if(curNode->left)
Q.push(curNode->left);
if(curNode->right)
Q.push(curNode->right);
count++;
}
if(level%2 ==0)     //偶数层，反转levNode再压入res
reverse(levNode.begin(),levNode.end());
res.push_back(levNode);
level++;
}
return res;
}
};

 思路二：

利用队列，在每一层结束时向栈中压入NULL， 则遇到NULL就标志一层的结束，就可以存节点，分奇偶的选择性的反转当前层的节点。特别值得注意的是，循环到最后一行后，若是还压入NULL，会造成死循环，故在压人入NULL时要判断栈是否为空。这也是在

层次遍历为主题的基础上增加层反转条件即可。

/**
* 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;
queue<TreeNode *> Q;
if(root==NULL)  return res;
Q.push(root);
Q.push(NULL);              //第一层的结束
vector<int> levNode;       //存放每层的节点
size_t level=1;            //第level层，初始为第一层
while( !Q.empty())
{
TreeNode *cur=Q.front();
Q.pop();
if(cur)
{
levNode.push_back(cur->val);    //必须在if内，因栈顶节点可能为NULL
if(cur->left)
Q.push(cur->left);
if(cur->right)
Q.push(cur->right);
}
else
{
if(level%2 ==0)        //偶数层，反转levNode再压入res
reverse(levNode.begin(),levNode.end());
res.push_back(levNode);
level++;

levNode.clear();
if( !Q.empty())   //最后一行，不要压入
Q.push(NULL);
}
}
return res;
}
};