420 sequence traversal of binary tree 2 (429. sequence traversal of n-ary tree, 515. find the maximum value in each tree row, 116. fill in the next right node pointer of each node, 104. maximum depth

429. N Sequence traversal of the fork tree

class Solution {
    
public:
    vector<vector<int>> levelOrder(Node* root) {
    
        vector<vector<int>> res;
        queue<Node*> que;

        if (root != nullptr) que.push(root);

        while (!que.empty())
        {
    
            vector<int> temp;
            int size = que.size();

            for (int i = 0; i < size; i++)
            {
    
                Node* node = que.front();
                que.pop();

                temp.push_back(node->val);

                for (int i = 0; i < (node->children.size()); i++)
                {
    
                    if (node->children[i]) que.push(node->children[i]);
                }
            }
            res.push_back(temp);
        }
        return res;
    }
};

515. Find the maximum in each tree row

class Solution {
    
public:
    vector<int> largestValues(TreeNode* root) {
    
        vector<int> res;
        queue<TreeNode*> que;

        if (root != nullptr) que.push(root);

        while (!que.empty())
        {
    
            int size = que.size();
            int temp = INT32_MIN;

            for (int i = 0; i < size; i++)
            {
    
                TreeNode* node = que.front();
                que.pop();

                temp = (node->val > temp) ? node->val : temp;
                if (node->left)  que.push(node->left);
                if (node->right) que.push(node->right);
            }
            res.push_back(temp);
        }
        return res;
    }
};

116. Fill in the next right node pointer for each node

class Solution {
    
public:
    Node* connect(Node* root) {
    
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
    
            int size = que.size();
            // vector<int> vec;
            Node* nodePre;
            Node* node;
            for (int i = 0; i < size; i++) {
    
                if (i == 0) {
    
                    nodePre = que.front(); //  Take out the head node of the first layer 
                    que.pop();
                    node = nodePre;
                } else {
    
                    node = que.front();
                    que.pop();
                    nodePre->next = node; //  Previous node of this layer next Point to this node 
                    nodePre = nodePre->next;
                }
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            nodePre->next = NULL; //  The last node of this layer points to NULL
        }
        return root;

    }
};

104. The maximum depth of a binary tree

class Solution {
    
public:
    int maxDepth(TreeNode* root) {
    
        queue<TreeNode*> que;
        int res = 0;
        if (root)    que.push(root);

        while (!que.empty())
        {
    
            int size = que.size();

            for (int i = 0; i < size; i++)
            {
    
                TreeNode* node = que.front();
                que.pop();

                if (node->left)  que.push(node->left);
                if (node->right)  que.push(node->right);
            }
            res++;
        }
        return res;
    }
};

111. Minimum depth of binary tree

class Solution {
    
public:
    int minDepth(TreeNode* root) {
    
        queue<TreeNode*> que;
        int res = 0;

        if (root)    que.push(root);

        while (!que.empty())
        {
    
            int size = que.size();
            res++;
            for (int i = 0; i < size; i++)
            {
    
                TreeNode* node = que.front();
                que.pop();

                if (node->left == nullptr && node->right == nullptr)   return res;
                if (node->left)  que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return res;
    }
};