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# 第六章 二叉树 part02 | ||
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今日内容: | ||
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● 层序遍历 10 | ||
● 226.翻转二叉树 | ||
● 101.对称二叉树 2 | ||
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详细布置 | ||
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## 层序遍历 | ||
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看完本篇可以一口气刷十道题,试一试, 层序遍历并不难,大家可以很快刷了十道题。 | ||
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题目链接/文章讲解/视频讲解:https://programmercarl.com/0102.%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E5%B1%82%E5%BA%8F%E9%81%8D%E5%8E%86.html | ||
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## 226.翻转二叉树 (优先掌握递归) | ||
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这道题目 一些做过的同学 理解的也不够深入,建议大家先看我的视频讲解,无论做过没做过,都会有很大收获。 | ||
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题目链接/文章讲解/视频讲解:https://programmercarl.com/0226.%E7%BF%BB%E8%BD%AC%E4%BA%8C%E5%8F%89%E6%A0%91.html | ||
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## 101. 对称二叉树 (优先掌握递归) | ||
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先看视频讲解,会更容易一些。 | ||
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题目链接/文章讲解/视频讲解:https://programmercarl.com/0101.%E5%AF%B9%E7%A7%B0%E4%BA%8C%E5%8F%89%E6%A0%91.html |
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# 101. 对称二叉树 | ||
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## 题目描述 | ||
## 解题思路 | ||
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```python | ||
# Definition for a binary tree node. | ||
# class TreeNode: | ||
# def __init__(self, val=0, left=None, right=None): | ||
# self.val = val | ||
# self.left = left | ||
# self.right = right | ||
class Solution: | ||
def f(self,a,b): | ||
if a == None and b != None: | ||
return False | ||
elif a == None and b == None: | ||
return True | ||
elif a != None and b == None: | ||
return False | ||
else: | ||
if a.val != b.val: | ||
return False | ||
else: | ||
res1 = self.f(a.left, b.right) | ||
res2 = self.f(a.right, b.left) | ||
return res1 and res2 | ||
def isSymmetric(self, root: Optional[TreeNode]) -> bool: | ||
if root == None: | ||
return True | ||
else: | ||
return self.f(root.left, root.right) | ||
``` | ||
## 学习感想 |
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# 102. 二叉树的层序遍历 | ||
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## 题目描述 | ||
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给你二叉树的根节点 root ,返回其节点值的 层序遍历 。 (即逐层地,从左到右访问所有节点)。 | ||
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## 解题思路 | ||
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```cpp | ||
class Solution { | ||
public: | ||
vector<vector<int>> levelOrder(TreeNode* root) { | ||
vector<vector<int>> res ; | ||
if (root == NULL) return res ; | ||
deque<TreeNode*> v; | ||
v.push_back(root); | ||
while (!v.empty()) { | ||
int size = v.size(); | ||
vector<int> level_res ; | ||
for (int i = 0; i < size; i ++) { | ||
TreeNode * ptr = v.front(); | ||
v.pop_front(); | ||
level_res.push_back(ptr->val); | ||
if (ptr->left) v.push_back(ptr->left); | ||
if (ptr->right) v.push_back(ptr->right); | ||
} | ||
res.push_back(level_res) ; | ||
} | ||
return res; | ||
} | ||
}; | ||
``` | ||
## 学习感想 | ||
### 107. 二叉树的层序遍历 II | ||
```cpp | ||
class Solution { | ||
public: | ||
vector<vector<int>> levelOrderBottom(TreeNode* root) { | ||
vector<vector<int>> res ; | ||
if (root == NULL) return res ; | ||
deque<TreeNode*> v; | ||
v.push_back(root); | ||
while (!v.empty()) { | ||
int size = v.size(); | ||
vector<int> level_res ; | ||
for (int i = 0; i < size; i ++) { | ||
TreeNode * ptr = v.front(); | ||
v.pop_front(); | ||
level_res.push_back(ptr->val); | ||
if (ptr->left) v.push_back(ptr->left); | ||
if (ptr->right) v.push_back(ptr->right); | ||
} | ||
res.push_back(level_res) ; | ||
} | ||
reverse(res.begin(),res.end()); | ||
return res; | ||
} | ||
}; | ||
``` | ||
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### 199. 二叉树的右视图 | ||
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```cpp | ||
class Solution { | ||
public: | ||
vector<int> rightSideView(TreeNode* root) { | ||
vector<int> res ; | ||
if (root == NULL) return res ; | ||
deque<TreeNode*> v; | ||
v.push_back(root); | ||
while (!v.empty()) { | ||
int size = v.size(); | ||
vector<int> level_res ; | ||
for (int i = 0; i < size; i ++) { | ||
TreeNode * ptr = v.front(); | ||
v.pop_front(); | ||
level_res.push_back(ptr->val); | ||
if (ptr->left) v.push_back(ptr->left); | ||
if (ptr->right) v.push_back(ptr->right); | ||
} | ||
res.push_back(level_res.back()); | ||
} | ||
return res; | ||
} | ||
}; | ||
``` | ||
### 637. 二叉树的层平均值 |
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# 226. 翻转二叉树 | ||
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## 题目描述 | ||
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给你一棵二叉树的根节点 root ,翻转这棵二叉树,并返回其根节点。 | ||
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## 解题思路 | ||
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```cpp | ||
class Solution { | ||
public: | ||
TreeNode* invertTree(TreeNode* root) { | ||
traverse(root); | ||
return root; | ||
} | ||
void traverse(TreeNode* root) { | ||
if (root == NULL) return ; | ||
swap(root->left, root->right); | ||
traverse(root->left); | ||
traverse(root->right); | ||
} | ||
}; | ||
``` | ||
## 学习感想 |