LeetCode算法题117：填充每个节点的下一个右侧节点指针 II解析

给定一个二叉树

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

这个题和上一个题基本思路还是一样的，递归的话需要判断左右节点是否存在，要链接到next的存在的一个节点，因此每次都要先连右边的，从后往前连。迭代的话需要设置一个空指针，因为可能存在左节点没有的情况，然后依次逐层连接即可。

C++源代码：

/*
// 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) {
        if (!root) return NULL;
        Node *p = root->next;
        while(p){
            if (p->left){
                p = p->left;
                break;
            }
            if(p->right){
                p = p->right;
                break;
            }
            p = p->next;
        }
        if (root->right) root->right->next = p;
        if (root->left) root->left->next = root->right?root->right:p;
        connect(root->right);
        connect(root->left);
        return root;
    }
};

python3源代码：

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        dummy = Node(0, None, None, None)
        cur = dummy
        p = root
        while p:
            if p.left:
                cur.next = p.left
                cur = cur.next
            if p.right:
                cur.next = p.right
                cur = cur.next
            p = p.next
            if p==None:
                p = dummy.next
                dummy.next = None
                cur = dummy
        return root