# 二叉树的锯齿形层序遍历
给定一个二叉树,返回其节点值的锯齿形层序遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
例如:
给定二叉树 [3,9,20,null,null,15,7]
,
3
/ \
9 20
/ \
15 7
返回锯齿形层序遍历如下:
[
[3],
[20,9],
[15,7]
]
以下错误的选项是?
## aop
### before
```c
#include
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```c
```
## 答案
```c
class Solution
{
public:
vector> zigzagLevelOrder(TreeNode *root)
{
vector> result;
if (root == NULL)
{
return result;
}
deque node_queue;
node_queue.push_back(root);
node_queue.push_back(NULL);
deque level_list;
bool is_order_left = true;
while (node_queue.size() > 0)
{
TreeNode *curr_node = node_queue.front();
node_queue.pop_front();
if (curr_node != NULL)
{
if (curr_node->left != NULL)
{
node_queue.push_back(curr_node->left);
}
if (curr_node->right != NULL)
{
node_queue.push_back(curr_node->right);
}
}
else
{
vector tmp;
for (int a : level_list)
{
tmp.push_back(a);
}
result.push_back(tmp);
level_list.clear();
if (node_queue.size() > 0)
{
node_queue.push_back(NULL);
}
is_order_left = !is_order_left;
}
}
return result;
}
};
```
## 选项
### A
```c
class Solution
{
public:
vector> zigzagLevelOrder(TreeNode *root)
{
if (root == NULL)
return {};
vector> res;
int flag = 1;
stack a, b;
a.push(root);
while (!a.empty() || !b.empty())
{
vector cur;
while (flag == 1 && !a.empty())
{
root = a.top();
cur.push_back(root->val);
a.pop();
if (root->left != NULL)
b.push(root->left);
if (root->right != NULL)
b.push(root->right);
}
while (flag == -1 && !b.empty())
{
root = b.top();
cur.push_back(root->val);
b.pop();
if (root->right != NULL)
a.push(root->right);
if (root->left != NULL)
a.push(root->left);
}
flag = -1 * flag;
res.push_back(cur);
}
return res;
}
};
```
### B
```c
class Solution
{
public:
vector> zigzagLevelOrder(TreeNode *root)
{
vector> ans;
if (!root)
return ans;
queue qnode;
bool orderByLeft = true;
qnode.push(root);
while (!qnode.empty())
{
int levelSize = qnode.size();
deque level;
for (int i = 0; i < levelSize; i++)
{
auto curNode = qnode.front();
qnode.pop();
if (orderByLeft)
{
level.push_back(curNode->val);
}
else
{
level.push_front(curNode->val);
}
if (curNode->left != NULL)
qnode.push(curNode->left);
if (curNode->right != NULL)
qnode.push(curNode->right);
}
orderByLeft = !orderByLeft;
vector curlevel{level.begin(), level.end()};
ans.push_back(curlevel);
}
return ans;
}
};
```
### C
```c
class Solution
{
public:
vector> zigzagLevelOrder(TreeNode *root)
{
if (!root)
return {};
vector> res;
queue q{{root}};
int cnt = 0;
while (!q.empty())
{
vector oneLevel;
for (int i = q.size(); i > 0; i--)
{
TreeNode *t = q.front();
q.pop();
oneLevel.push_back(t->val);
if (t->left)
q.push(t->left);
if (t->right)
q.push(t->right);
}
if (cnt % 2 == 1)
reverse(oneLevel.begin(), oneLevel.end());
res.push_back(oneLevel);
cnt++;
}
return res;
}
};
```
### D
```c
class Solution
{
public:
vector> zigzagLevelOrder(TreeNode *root)
{
vector> results;
if (root == NULL)
{
return results;
}
zigzagLevelOrderDFS_helper(root, 0, results);
return results;
}
void zigzagLevelOrderDFS_helper(TreeNode *node, int level, vector> &results)
{
if (level >= results.size())
{
vector new_level;
new_level.push_back(node->val);
results.push_back(new_level);
}
else
{
if (level % 2 == 0)
{
results[level].push_back(node->val);
}
else
{
results[level].insert(results[level].begin(), node->val);
}
}
if (node->left != NULL)
{
zigzagLevelOrderDFS_helper(node->left, level + 1, results);
}
if (node->right != NULL)
{
zigzagLevelOrderDFS_helper(node->right, level + 1, results);
}
}
};
```