【Leetcode】116. Populating Next Right Pointers in Each Node（队列）

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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.

Example:

Input: {"$id":"1","left":{"$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}

Output: {"$id":"1","left":{"$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}

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

题目大意：

 给出一个完美二叉树。将同一水平线上的节点连接起来作为节点的next。

解题思路：

使用队列完美解决。

/*
// 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==NULL) return root;
        deque<Node*> c;
        
        c.push_back(root);
        
        while(c.size()){
            int cor_nums = c.size();
            Node* tmp = c.front();
            c.pop_front();
            if(tmp->left&&tmp->right){
                c.push_back(tmp->left);
                c.push_back(tmp->right);
            }
            for(int i=1;i<cor_nums;i++){
                Node* new_node = c.front();
                c.pop_front();
                
                if(new_node->left&&new_node->right){
                    c.push_back(new_node->left);
                    c.push_back(new_node->right);
                }
                tmp->next = new_node;
                tmp = tmp->next;
            }
            tmp->next=nullptr;
        }
        return root;
    }
};