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# 第六章 二叉树part01 | ||
今日内容: | ||
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● 理论基础 | ||
● 递归遍历 | ||
● 迭代遍历 | ||
● 统一迭代 | ||
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详细布置 | ||
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## 理论基础 | ||
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需要了解 二叉树的种类,存储方式,遍历方式 以及二叉树的定义 | ||
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文章讲解:https://programmercarl.com/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%90%86%E8%AE%BA%E5%9F%BA%E7%A1%80.html | ||
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## 递归遍历 (必须掌握) | ||
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二叉树的三种递归遍历掌握其规律后,其实很简单 | ||
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题目链接/文章讲解/视频讲解:https://programmercarl.com/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E9%80%92%E5%BD%92%E9%81%8D%E5%8E%86.html | ||
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## 迭代遍历 (基础不好的录友,迭代法可以放过) | ||
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题目链接/文章讲解/视频讲解:https://programmercarl.com/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E8%BF%AD%E4%BB%A3%E9%81%8D%E5%8E%86.html | ||
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## 统一迭代 (基础不好的录友,迭代法可以放过) | ||
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这是统一迭代法的写法, 如果学有余力,可以掌握一下 | ||
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题目链接/文章讲解:https://programmercarl.com/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E7%BB%9F%E4%B8%80%E8%BF%AD%E4%BB%A3%E6%B3%95.html |
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# 144. 二叉树的前序遍历 | ||
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## 题目描述 | ||
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## 解题思路 | ||
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```cpp | ||
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/** | ||
* Definition for a binary tree node. | ||
* 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) {} | ||
* }; | ||
*/ | ||
class Solution { | ||
public: | ||
vector<int> preorderTraversal(TreeNode* root) { | ||
vector<int> res ; | ||
traverse(root, res); | ||
return res ; | ||
} | ||
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void traverse(TreeNode* root, vector<int>& vec) { | ||
if (root == NULL) return ; | ||
vec.push_back(root->val); | ||
traverse(root->left, vec); | ||
traverse(root->right, vec); | ||
} | ||
}; | ||
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``` | ||
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递归的实现就是:每一次递归调用都会把函数的局部变量、参数值和返回地址等压入调用栈中 | ||
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递归的实质就是迭代 | ||
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统一格式法 | ||
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```cpp | ||
class Solution { | ||
public: | ||
vector<int> postorderTraversal(TreeNode* root) { | ||
vector<int> res ; | ||
vector<TreeNode *> s ; | ||
if (root != NULL) s.push_back(root); | ||
while (!s.empty()) { | ||
TreeNode* last = s.back(); | ||
s.pop_back(); | ||
if (last == NULL) { | ||
TreeNode* last = s.back(); | ||
s.pop_back(); | ||
res.push_back(last->val); | ||
} else { | ||
s.push_back(last); | ||
s.push_back(NULL); | ||
if (last->right) s.push_back(last->right); | ||
if (last->left) s.push_back(last->left); | ||
} | ||
} | ||
return res ; | ||
} | ||
}; | ||
``` | ||
## 学习感想 |