1.leetCode 100. 相同的树

C++代码：

class Solution {
public:bool isSameTree(TreeNode* p, TreeNode* q) {if(p == nullptr || q == nullptr) return p==q;return p->val == q->val && isSameTree(p->left,q->left) && isSameTree(p->right,q->right);}
};

Python代码： 

class Solution:def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:if p is None or q is None:return p is qreturn p.val == q.val and self.isSameTree(p.left,q.left) and self.isSameTree(p.right,q.right)

2.leetCode 101. 对称二叉树
力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

C++代码： 

class Solution {
public:bool isSameTree(TreeNode* p,TreeNode* q) {if(p == nullptr || q == nullptr) return p==q;return p->val == q->val && isSameTree(p->left,q->right) && isSameTree(p->right,q->left);}bool isSymmetric(TreeNode* root) {if(root==nullptr) return true;TreeNode* p = root->left;TreeNode* q = root->right;return isSameTree(p,q);}
};

 Python代码： 

class Solution:def isSameTree(self,p:Optional[TreeNode],q:Optional[TreeNode]) -> bool:if p is None or q is None:return p is qreturn p.val == q.val and self.isSameTree(p.left,q.right) and self.isSameTree(p.right,q.left)def isSymmetric(self, root: Optional[TreeNode]) -> bool:return self.isSameTree(root.left,root.right)

3.leetCode 110.平衡二叉树
https://leetcode.cn/problems/balanced-binary-tree/description/

C++代码：

class Solution {
public:int getHeight(TreeNode* root) {if(root==nullptr) return 0;int LH = getHeight(root->left);if(LH == -1) return -1;int RH = getHeight(root->right);if(RH == -1 or abs(RH-LH) > 1) return -1;return max(LH,RH)+1;}bool isBalanced(TreeNode* root) {return getHeight(root)!=-1;}
};

Python代码：

class Solution:def isBalanced(self, root: Optional[TreeNode]) -> bool:def get_height(node):if node is None: return 0LH = get_height(node.left)if LH == -1:return -1RH = get_height(node.right)if RH == -1 or abs(RH-LH) > 1:return -1return max(LH,RH)+1return get_height(root)!=-1

4.leetCode 199. 二叉树的右视图
https://leetcode.cn/problems/binary-tree-right-side-view/description/

C++代码： 

class Solution {
public:vector<int> ans;void dfs(TreeNode* node,int depth) {if(node==nullptr) return;if(depth == ans.size()) ans.push_back(node->val);dfs(node->right,depth+1);dfs(node->left,depth+1);}vector<int> rightSideView(TreeNode* root) {dfs(root,0);return ans;}
};

Python代码： 

class Solution:def rightSideView(self, root: Optional[TreeNode]) -> List[int]:ans = []def f(node,depth):if node is None:return if depth == len(ans):ans.append(node.val)f(node.right,depth+1)f(node.left,depth+1)f(root,0)return ans

参考和推荐文章、视频：

如何灵活运用递归？【基础算法精讲 10】_哔哩哔哩_bilibili

100. 相同的树 https://leetcode.cn/problems/same-tree/solutions/2015056/ru-he-ling-huo-yun-yong-di-gui-lai-kan-s-empk/

101. 对称二叉树 https://leetcode.cn/problems/symmetric-tree/solutions/2015063/ru-he-ling-huo-yun-yong-di-gui-lai-kan-s-6dq5/

110. 平衡二叉树 https://leetcode.cn/problems/balanced-binary-tree/solutions/2015068/ru-he-ling-huo-yun-yong-di-gui-lai-kan-s-c3wj/

199. 二叉树的右视图 https://leetcode.cn/problems/binary-tree-right-side-view/solutions/2015061/ru-he-ling-huo-yun-yong-di-gui-lai-kan-s-r1nc/