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一、计算布尔二叉树的值

1、题目描述

leetcode链接

2、代码

class Solution 
{
public:bool evaluateTree(TreeNode* root) {// 出口if(root->left == nullptr){if(root->val == 0){return false;}else{return true;}}// 左子树bool Left = evaluateTree(root->left);bool Right = evaluateTree(root->right);// 结点往上返回if(root->val == 2){return Left | Right;}else{return Left & Right;}}
};

3、解析

二、求根节点到叶节点数字之和

1、题目描述

leetcode链接

2、代码

class Solution 
{
public:int sumNumbers(TreeNode* root) {return dfs(root, 0);}int dfs(TreeNode* root, int persum){persum = persum * 10 + root->val;// 1、出口if(root->left == nullptr && root->right == nullptr){return persum;}// 2、函数体int ret = 0;if(root->left){ret += dfs(root->left, persum);}if(root->right){ret += dfs(root->right, persum);}return ret;}
};

3、解析

三、二叉树剪枝

1、题目描述

leetcode链接

2、代码

class Solution 
{
public:TreeNode* pruneTree(TreeNode* root) {// 递归出口if(root == nullptr){return nullptr;}// 处理左右子树root->left = pruneTree(root->left);root->right = pruneTree(root->right);// 判断if(root->left == nullptr && root->right == nullptr && root->val == 0){delete root;root = nullptr;}return root;}
};

3、解析

四、验证二叉搜索树

1、题目描述

leetcode链接

2、代码

class Solution 
{
public:// 定义一个prev值为最小long prev = LONG_MIN;bool isValidBST(TreeNode* root) {// 出口if(root == nullptr){return true;}bool Lefttree = isValidBST(root->left);// 剪枝if(Lefttree == false)return false;bool cur = false; // 判断当前是不是二叉搜索树if(root->val > prev){cur = true;}// 剪枝if(cur == false){return false;}prev = root->val; // 更新bool Righttree = isValidBST(root->right);if(Lefttree == true && Righttree == true && cur == true){return true;}else{return false;}}
};

3、解析

五、二叉搜索树中第K小的元素

1、题目描述

leetcode链接

2、代码

class Solution 
{
public:int ret = 0;int count = 0;int kthSmallest(TreeNode* root, int k) {count = k;dfs(root);return ret;}void dfs(TreeNode* root){// 1、出口if(root == nullptr || count == 0){return;}// 2、中序遍历dfs(root->left);count--;if(count == 0){// 3、优化：剪枝ret = root->val;return;}dfs(root->right);}
};

3、解析

六、二叉树的所有路径

1、题目描述

leetcode链接

2、代码

// 不加剪枝
class Solution 
{
public:// 1、定义一个全局返回变量retvector<string> ret;vector<string> binaryTreePaths(TreeNode* root) {// 2、进行dfs函数的调用string path;dfs(root, path);return ret;}void dfs(TreeNode* root, string path){// 3、递归出口if(root == nullptr) return;// 4、叶子结点的增加路径if(root->right == nullptr && root->left == nullptr){path += to_string(root->val);ret.push_back(path);}// 5、非叶子节点path += to_string(root->val);path += "->";// 6、递归dfs(root->left, path);dfs(root->right, path);}
};

// 加剪枝
class Solution 
{
public:vector<string> ret;vector<string> binaryTreePaths(TreeNode* root) {string path;dfs(root, path);return ret;}void dfs(TreeNode* root, string path){if(root->left == nullptr && root->right == nullptr){path += to_string(root->val);ret.push_back(path);return;}path += to_string(root->val);path += "->";if(root->left) dfs(root->left, path);if(root->right) dfs(root->right, path);}
};

3、解析