题目描述: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]
]
思路分析:
- 判空,若该二叉树为空,直接返回;
- 不为空时:(判断不为空之前插入该二叉树的根节点)
1.创建一个队列,将根节点放入队列中,并且创建一个容器,将队列中的数据放入容器中,同时将队列中的根节点删掉;
2.接着取该二叉树的子节点,当左右子节点不为空时,压入队列中;
3.此时需要对偶数层节点进行翻转(在此之间定义一个计数器,如果计数器定义为0,则判断翻转的条件是:if(count%2!=0
如果计数器定义为1,则判断翻转的条件是:if(count%2==0
)
代码实现如下
class Solution {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int> >res;
int count =1;//定义计数器为1,注意判断翻转的条件
if(root==NULL)//判空
return res;
queue<TreeNode*>q;
q.push(root);//压入根节点
while(!q.empty())
{
int n = q.size();
vector<int> vec;
while(vec.size() < n)
{
TreeNode* root = q.front();//对头元素为根节点
q.pop();
vec.push_back(root->val);
if(root->left)
{
q.push(root->left);
}
if(root->right)
{
q.push(root->right);
}
}
count++;
if(count%2==0)
{
reverse(vec.begin(), vec.end());
}
res.push_back(vec);
}
return res;
}
};