117. Populating Next Right Pointers in Each Node II-优快云博客

本文探讨了如何在任意二叉树中填充每个节点的Next指针，使其指向同一层级的下一个节点。解决方案需满足仅使用常数额外空间的要求，并提供了两种不同的实现方法。

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Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

After calling your function, the tree should look like:

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

void connect(TreeLinkNode *root) {
   if (!root) return;
    TreeLinkNode dummy(INT_MIN);
    for (TreeLinkNode *cur = root, *pre = &dummy; cur; cur = cur->next) {
        if (cur->left) {
            pre->next = cur->left;
            pre = pre->next;
        }
        if (cur->right) {
            pre->next = cur->right;
            pre = pre->next;
        }
    }
    connect(dummy.next);
}

void connect(TreeLinkNode *root) {
    TreeLinkNode *now, *tail, *head;
    
    now = root;
    head = tail = NULL;
    while(now)
    {
        if (now->left)
            if (tail) tail = tail->next =now->left;
            else head = tail = now->left;
        if (now->right)
            if (tail) tail = tail->next =now->right;
            else head = tail = now->right;
        if(!(now = now->next))
        {
            now = head;
            head = tail=NULL;
        }
    }
}