南京哪里有做公司网站的集合竞价口诀背熟6句

222.完全二叉树的节点个数

题目链接
文章讲解

/*** 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 countNodes(TreeNode* root) {queue<TreeNode*> q;if(root) q.push(root);vector<int> res;int ans=0;while(!q.empty()){int k=q.size();ans+=k;while(k--){TreeNode* node=q.front();q.pop();if(node->left) q.push(node->left);if(node->right) q.push(node->right);}}return ans;}
};

110.平衡二叉树

题目链接
文章讲解

/*** 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 solve(TreeNode* cur){if(cur==NULL) return 0;int left=solve(cur->left);if(left==-1) return -1;int right=solve(cur->right);if(right==-1) return -1;if(abs(left-right)>1) return -1;else return max(left,right)+1;}bool isBalanced(TreeNode* root) {if(solve(root)==-1) return false;else return true;}
};

257. 二叉树的所有路径

题目链接
文章讲解

/*** 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 {
private:void traversal(TreeNode* cur, string path, vector<string>& result) {path += to_string(cur->val); // 中if (cur->left == NULL && cur->right == NULL) {result.push_back(path);return;}if (cur->left) traversal(cur->left, path + "->", result); // 左if (cur->right) traversal(cur->right, path + "->", result); // 右}public:vector<string> binaryTreePaths(TreeNode* root) {vector<string> result;string path;if (root == NULL) return result;traversal(root, path, result);return result;}
};

404.左叶子之和

题目链接
文章讲解

/*** 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 sumOfLeftLeaves(TreeNode* root) {if(root->left==NULL&&root->right==NULL) return 0;queue<TreeNode*> q;if(root) q.push(root);vector<int> res;int ans=0;while(!q.empty()){int k=q.size();for(int i=1;i<=k;i++){TreeNode* node=q.front();q.pop();if(node->left) if(node->left->left==NULL&&node->left->right==NULL) ans+=node->left->val; else q.push(node->left);if(node->right) q.push(node->right);}}return ans;}
};