leetcode116 - Populating Next Right Pointers in Each Node -medium

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 = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
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. 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 4096.
• -1000 <= node.val <= 1000

 

层序遍历的思想。每一层记录首节点first，对于当前节点，三步1. check if left，cur->left->next = cur->right 2. check if right&next, cur->right->next = cur->next->left 3. move cur to cur->next. 当前这层搞完，move去下一层的首节点，即first->left. 好处是每处理到一个节点，可以利用到它的next指针一定已经处理好了，要么null要么已经指好了。

 

实现：Time O(n) Space O(1)

class Solution {
public:
    Node* connect(Node* root) {
        if (!root) return nullptr;
        Node* first = root;
        while (first){
            Node* cur = first;
            while (cur){ //same level
                if (cur->left)
                    cur->left->next = cur->right;
                if (cur->right && cur->next)
                    cur->right->next = cur->next->left;
                cur = cur->next;
            }
            first = first->left;
        }
        return root;
    }
};