[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.
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}".
/**
* 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=; //第level层,初始为第一层
while( !Q.empty())
{
int count=;
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% ==) //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=; //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% ==) //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;
}
};
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