Leetcode Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *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.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

 

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

 

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

此題迭代即可，注意在迭代下一行時，前一行已經迭代完，可以通過node->next指標去訪問下一個節點，
如果node->next到達末尾，則下一行的next已經新增完畢，則可以切換下一行遍歷（通過上一行的第一個節點的left切換）
相當於遍歷二維陣列

void connect(TreeLinkNode *root){
    if(root == NULL) return;
    root->next = NULL;
    TreeLinkNode *p = root;
    while(p){
        TreeLinkNode *q = p;
        while(q){
            if(q->left) q->left->next = q->right;
            if(q->right && q->next) q->right->next = q->next->left;
            q=q->next;
        }
        p = p->left;
    }
}