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;
}
}

};