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

此题迭代即可，注意在迭代下一行时，前一行已经迭代完，可以通过node->next指针去访问下一个节点，
如果node->next到达末尾，则下一行的next已经添加完毕，则可以切换下一行遍历（通过上一行的第一个节点的left切换）
相当于遍历二维数组

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;
    }
}