117. Populating Next Right Pointers in Each Node II**

117. Populating Next Right Pointers in Each Node II**

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

题目描述

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Follow up:

• You may only use constant extra space.
• Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

Example 1:

Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Constraints:

• The number of nodes in the given tree is less than 6000.
• -1000 <= node.val <= 1000

C++ 实现 1

此题是 116. Populating Next Right Pointers in Each Node** 这道题的升级. 但是发现我用层序遍历的代码不用做任何修改就可以用于此题!

这道题使用层序遍历应该是非常方便的, 使用 prev 表示前一个指针, 然后让每一层中的节点的 next 指向 prev 并不断更新 prev. 注意先右孩子如队列, 再左孩子进队列.

/*
// 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;
        q.push(root);
        while (!q.empty()) {
            auto size = q.size();
            Node *prev = nullptr;
            while (size --) {
                auto r = q.front();
                q.pop();
                r->next = prev;
                prev = r;
                if (r->right) q.push(r->right);
                if (r->left) q.push(r->left);
            }
        }
        return root;
    }
};