LeetCode117. Populating Next Right Pointers in Each Node II

一、题目

Given a binary tree

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 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.
Example 2:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 6000].
-100 <= Node.val <= 100

二、题解

与上题思路完全一致，利用层序遍历

/*
// 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) {
        queue<Node*> q;
        if(root != NULL) q.push(root);
        while(!q.empty()){
            int size = q.size();
            Node* pre = NULL;
            while(size--){
                Node* cur = q.front();
                q.pop();
                if(cur->left != NULL) q.push(cur->left);
                if(cur->right != NULL) q.push(cur->right);
                if(pre != NULL) pre->next = cur;
                pre = cur;
                if(size == 0) cur->next = NULL;
            }
        }
        return root;
    }
};