1.计算布尔二叉树的值

1.1 题目

题目链接

1.2 思路

1.3 代码

/*** 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:bool evaluateTree(TreeNode* root) {if(root->left == nullptr) return root->val == 0 ? false : true;bool left = evaluateTree(root->left);bool right = evaluateTree(root->right);return root->val == 2 ? left || right : left && right;}
};

2.求根节点到叶节点数字之和

2.1 题目

题目链接

2.2 思路

2.3 代码

/*** 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:int sumNumbers(TreeNode* root) {return dfs(root, 0);}int dfs(TreeNode* root, int presum){presum = presum * 10 + root->val;if(root->left == nullptr && root->right == nullptr) return presum;int ret = 0;if(root->left) ret += dfs(root->left, presum);if(root->right) ret += dfs(root->right, presum);return ret;}
};

3.二叉树剪枝

3.1 题目

题目链接

3.2 思路

3.3 代码

/*** 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:TreeNode* pruneTree(TreeNode* root) {if(root == nullptr) return nullptr;root->left = pruneTree(root->left);root->right = pruneTree(root->right);if(root->left == nullptr && root->right == nullptr && root->val == 0)root = nullptr;return root;}
};

4.验证二叉搜索树

4.1 题目

题目链接

4.2 思路

4.3 代码

/*** 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 {long prev = LONG_MIN;
public:bool isValidBST(TreeNode* root) {if(root == nullptr) return true;bool left = isValidBST(root->left);// 剪枝if(left == false) return false;if(root->val <= prev) return false;prev = root->val;bool right = isValidBST(root->right);return left && right; }
};

5.二叉搜索树中第K小的元素

5.1 题目

题目链接

5.2 思路

5.3 代码

/*** 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 {int ret, count;
public:int kthSmallest(TreeNode* root, int k) {count = k;dfs(root);return ret;}void dfs(TreeNode* root){if(root == nullptr || count == 0) return;dfs(root->left);count--;if(count == 0) ret = root->val;dfs(root->right);}
};

6.二叉树的所有路径

6.1 题目

题目链接

6.2 思路

6.3 代码

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