【树】

题目：

 

 

解法：

 

 

 

/*
// 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 == NULL)
        {
            return root;
        }

        Node *leftwall = root;
        while(leftwall != NULL)
        {
            Node *across = leftwall;
            while(across != NULL)
            {
                //如果左子树不为空，则设置其左子树的next域
                if(across->left != NULL)
                {
                    across->left->next = across->right;
                }

                //如果右子树不为空，且该右子树的叔叔结点的左孩子不为空
                if(across->right != NULL && across->next != NULL)
                {
                    across->right->next = across->next->left;
                }
                across = across->next;
            }

            leftwall = leftwall->left;
        }

        return root;
    }
};