117. Populating Next Right Pointers in Each Node II

Solution - Recursion

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root)
    {
        if (!root) return nullptr;
        // 使用一個指標指向下一個節點
        Node* cur = root->next;
        
        while (cur)
        {
            if (cur->left)
            {
                cur = cur->left;
                break;
            }
            if (cur->right)
            {
                cur = cur->right;
                break;
            }
            cur = cur->next;
        }

        if (root->right)
        {
            root->right->next = cur;
        }
        if (root->left)
        {
            root->left->next = root->right ? root->right : cur;
        }
        connect(root->right);
        connect(root->left);
        return root;        
    }
};

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(n)$

Solution - Iteration

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root)
    {
        if (!root) return nullptr;
        
        queue<Node*> q{{root}};
    
        while (!q.empty())
        {
            int qSize = q.size();
            for (int i = 0; i < qSize; i++)
            {
                Node* node = q.front(); q.pop();
                if (i < qSize - 1)
                {
                    // 如果不是最末端的 node，node->next 指到下一個節點
                    node->next = q.front();
                }

                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
                
            }
        }
        return root;
    }
};

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(n)$

Solution - Space O(1)

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        Node* dummy = new Node(-1);
        Node* cur = dummy;
        Node* head = root;

        while (root)
        {
            if (root->left)
            {
                cur->next = root->left;
                cur = cur->next;
            }
            if (root->right)
            {
                cur->next = root->right;
                cur = cur->next;
            }

            root = root->next;
            
            if (!root)
            {
                cur = dummy;
                root = dummy->next;
                dummy->next = NULL;
            }
        }
        return head;
    }
};

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(1)$