116. 填充每个节点的下一个右侧节点指针

题目链接：leetcode.
层次遍历

/*
// 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 == nullptr)
            return root;
        queue<Node*> Q;
        Q.push(root);
        int cnt = 0;
        while(!Q.empty())
        {
            int t = Q.size();
            while(t--)
            {
                Node* tmp = Q.front();
                Q.pop();
                if(t)
                {
                    tmp -> next = Q.front();
                }
                if(tmp -> left)
                {
                    Q.push(tmp -> left);
                }
                if(tmp -> right)
                {
                    Q.push(tmp -> right);
                }
            }
        }
        return root;
    }
};

利用上一层与下一层之间的关系原地变化

/*
执行用时：20 ms, 在所有 C++ 提交中击败了86.44%的用户
内存消耗：16.4 MB, 在所有 C++ 提交中击败了84.94%的用户
*/
/*
// 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 == nullptr)
            return root;
        if(root -> left)
            root -> left -> next = root -> right;
        if(root -> next && root -> right)
        {
            root -> right -> next = root -> next -> left;
        }
        connect(root -> left);
        connect(root -> right);
        return root;
    }
};