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youngyangyang04
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|[0053.最大子序和](https://github.com/youngyangyang04/leetcode/blob/master/problems/0053.最大子序和.md) |数组 |简单|**暴力** **贪心** 动态规划 分治| |[0053.最大子序和](https://github.com/youngyangyang04/leetcode/blob/master/problems/0053.最大子序和.md) |数组 |简单|**暴力** **贪心** 动态规划 分治|
|[0059.螺旋矩阵II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0059.螺旋矩阵II.md) |数组 |中等|**模拟**| |[0059.螺旋矩阵II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0059.螺旋矩阵II.md) |数组 |中等|**模拟**|
|[0083.删除排序链表中的重复元素](https://github.com/youngyangyang04/leetcode/blob/master/problems/0083.删除排序链表中的重复元素.md) |链表 |简单|**模拟**| |[0083.删除排序链表中的重复元素](https://github.com/youngyangyang04/leetcode/blob/master/problems/0083.删除排序链表中的重复元素.md) |链表 |简单|**模拟**|
|[0094.二叉树的中序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0094.二叉树的中序遍历.md) |树 |中等|**递归** **栈**|
|[0142.环形链表II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0142.环形链表II.md) |链表 |中等|**快慢指针/双指针**| |[0142.环形链表II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0142.环形链表II.md) |链表 |中等|**快慢指针/双指针**|
|[0144.二叉树的前序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0144.二叉树的前序遍历.md) |树 |中等|**递归** **栈**|
|[0151.翻转字符串里的单词](https://github.com/youngyangyang04/leetcode/blob/master/problems/0151.翻转字符串里的单词.md) |字符串 |中等|**模拟/双指针**| |[0151.翻转字符串里的单词](https://github.com/youngyangyang04/leetcode/blob/master/problems/0151.翻转字符串里的单词.md) |字符串 |中等|**模拟/双指针**|
|[0155.最小栈](https://github.com/youngyangyang04/leetcode/blob/master/problems/0155.最小栈.md) |栈 |简单|**栈**| |[0155.最小栈](https://github.com/youngyangyang04/leetcode/blob/master/problems/0155.最小栈.md) |栈 |简单|**栈**|
|[0202.快乐数](https://github.com/youngyangyang04/leetcode/blob/master/problems/0202.快乐数.md) |哈希表 |简单|**哈希**| |[0202.快乐数](https://github.com/youngyangyang04/leetcode/blob/master/problems/0202.快乐数.md) |哈希表 |简单|**哈希**|

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## 题目地址
https://leetcode-cn.com/problems/binary-tree-inorder-traversal/
## 思路
## C++代码
递归
```
class Solution {
public:
void traversal(TreeNode* root, vector<int>& vec) {
if (root == NULL) return;
traversal(root->left, vec);
vec.push_back(root->val);
traversal(root->right, vec);
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result;
traversal(root, result);
return result;
}
};
```
> 更多算法干货文章持续更新可以微信搜索「代码随想录」第一时间围观关注后回复「Java」「C++」 「python」「简历模板」「数据结构与算法」等等就可以获得我多年整理的学习资料。

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## 题目地址
https://leetcode-cn.com/problems/binary-tree-preorder-traversal/
## 思路
实现树的遍历有使用递归和使用栈两种思路
## C++代码
递归
```
class Solution {
public:
void traversal(TreeNode* root, vector<int>& vec) {
if (root == NULL) return;
vec.push_back(root->val);
traversal(root->left, vec);
traversal(root->right, vec);
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> result;
traversal(root, result);
return result;
}
};
```
```
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
stack<TreeNode*> st;
vector<int> result;
st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
st.pop();
if (node != NULL) result.push_back(node->val);
else continue;
st.push(node->right);
st.push(node->left);
}
return result;
}
};
```
> 更过算法干货文章持续更新可以微信搜索「代码随想录」第一时间围观关注后回复「Java」「C++」 「python」「简历模板」「数据结构与算法」等等就可以获得我多年整理的学习资料。