LeetCode:Populating Next Right Pointers in Each Node

问题描述：

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

思路：首先想到的是BFS。但是BFS需要借助队列，这样就不能满足题目空间复杂度的要求。难点在于如何不用多余的空间完成树的遍历。解决这个问题需要用到树节点结构中包含的next域。可以借助next完成不用队列的BFS。

代码：

void Solution::connect(TreeLinkNode * root){    if(root == NULL)        return;    TreeLinkNode * cur;    while(root->left != NULL)//当root->left为NULL时，表示已经遍历到了最后一层，此时已经完成了所有next的赋值    {        cur = root;//在遍历每一层时，cur指针用来指向当前所要处理的节点        while(cur != NULL)//cur为NULL时，表示当前这一层已经处理完        {            if(cur->left != NULL)                cur->left->next = cur->right;//当前节点左子节点的next域指向当前节点的右子节点            if(cur->next != NULL)                cur->right->next = cur->next->left;//当前节点的右子节点的next域指向当前节点next域所指节点的左子节点            cur = cur->next;        }        root = root->left;//开始处理下一层    }}