NO.116 填充每个节点的下一个右侧节点指针

给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

 

示例：

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

 

提示：

    你只能使用常量级额外空间。
    使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。

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

    Node() {}

    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* first=root;
        if(root)root->next=NULL;
        while(first&&first->left)
        {
            Node* tmp=NULL;
            Node* next_first=first->left;
            while(first)
            {
                if(!tmp)
                {
                    tmp=first->left;
                    tmp->next=first->right;
                    tmp=tmp->next;
                    tmp->next=NULL;
                }
                else
                {
                    tmp->next=first->left;
                    tmp=tmp->next;
                    tmp->next=first->right;
                    tmp=tmp->next;
                    tmp->next=NULL;
                }
                first=first->next;
            }
            first=next_first;
        }
        return root;
    }
};

执行用时 :84 ms, 在所有C++提交中击败了93.78% 的用户

内存消耗 :27.3 MB, 在所有C++提交中击败了19.62%的用户