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

第一眼觉得，诶，用层序遍历不完了吗，但一看只能用常量空间来做，用队列来层序遍历肯定不行的。

还是想了好一会才想到，我们的构造是从上往下的，这一层连结好了，我们当前节点为 pre,那它的兄弟不就是pre->next吗，所以下一层的连接需要两步，一步是pre的左孩子连到右孩子去，一步是pre的右孩子连到 pre->next->left左孩子去，然后依次对pre的兄弟节点处理就完了。

代码：

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode* root)
{
    while(root)
	{
		TreeLinkNode* pre=root;
        TreeLinkNode* sib;
		while(pre)
		{
			if(pre->left)
				pre->left->next=pre->right;
			sib=pre->next;
			if ( sib && pre->right)
				pre->right->next=sib->left;
			pre=sib;
		}
		root=root->left;
	}
}

};