代码随想录算法训练营第1天|704. 二分查找、27. 移除元素

107. 二叉树的层序遍历II

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class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        queue<TreeNode*> que;
        if(root == nullptr)
            return result;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--)
            {
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node -> val);
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
            }
        result.push_back(vec);   
        }
        reverse(result.begin(), result.end());
        return result;
    }
};

199.二叉树的右视图

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class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> result;
        queue<TreeNode*> que;
        if(root == nullptr)
            return result;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            while(size--)
            {
                TreeNode* node = que.front();
                que.pop();
                if(size == 0)
                    result.push_back(node -> val);
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
            }   
        }
        //reverse(result.begin(), result.end());
        return result;
    }
};

637.二叉树的层平均值

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class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        vector<double> result;
        queue<TreeNode*> que;
        if(root == nullptr)
            return result;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--)
            {
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node -> val);
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
            }
        double sum = 0;
        for(auto num:vec)
            sum += num;
        double curRes = sum / (double) vec.size();
        result.push_back(curRes);   
        }
        return result;
    }
};

429.N叉树的层序遍历

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class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> result;
        queue<vector<Node*>> que;
        if(root == nullptr)
            return result;
        que.push(vector<Node*>{root});
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--)
            {
                vector<Node*> node = que.front();
                que.pop();
                int sizeNode = node.size();
                for(int i = 0; i <sizeNode; i++)
                {
                    vec.push_back(node[i] -> val);
                    if(node[i] -> children.size() != 0)
                        que.push(node[i] -> children);
                }
            }
            result.push_back(vec);   
        }
        return result;
    }
};

515.在每个树行中找最大值

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class Solution {
public:
    vector<int> largestValues(TreeNode* root) {
        vector<int> result;
        queue<TreeNode*> que;
        if(root == nullptr)
            return result;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--)
            {
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node -> val);
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
            }
        int max = vec[0];
        for(int i = 1; i < vec.size(); i++)
            if(vec[i] >max)
                max = vec[i];
        result.push_back(max);   
        }
        return result;
    }
};

116.填充每个节点的下一个右侧节点指针

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class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if(root == nullptr)
            return root;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--) 
            {
                Node* node = que.front();
                que.pop();
                if(size == 0)
                    node -> next = NULL;
                else 
                    node -> next = que.front(); 
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
            } 
        }
        return root; 
    }
};

117.填充每个节点的下一个右侧节点指针II

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class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if(root == nullptr)
            return root;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--) 
            {
                Node* node = que.front();
                que.pop();
                if(size == 0)
                    node -> next = NULL;
                else 
                    node -> next = que.front(); 
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
            } 
        }
        return root;
    }
};

104.二叉树的最大深度

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class Solution {
public:
    int maxDepth(TreeNode* root) {
        int depth = 0;
        queue<TreeNode*> que;
        if(root == nullptr)
            return 0;
        que.push(root);
        while(!que.empty())
        {
            depth++;
            int size = que.size();
            while(size--)
            {
                TreeNode* node = que.front();
                que.pop();
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
            }
        }
        return depth;
    }
};

111.二叉树的最小深度

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class Solution {
public:
    int minDepth(TreeNode* root) {
        int depth = 0;
        queue<TreeNode*> que;
        if(root == nullptr)
            return depth;
        que.push(root);
        while(!que.empty())
        {
            depth++;
            int size = que.size();
            while(size--)
            {
                TreeNode* node = que.front();
                que.pop();
                if(node -> left != nullptr)
                    que.push(node -> left);
                if(node -> right != nullptr)
                    que.push(node -> right);
                if(node -> left == nullptr && node -> right == nullptr)
                    return depth;
            }
        }
        return depth;
    }
};

总结

                     日期： 2023 年 3 月 29 日
              学习时长： 1 h 30 m
                     难度：$★$
累计完成题目数量： 47
距离目标还有数量： 253
                      小结：