Populating Next Right Pointers in Each Node -- leetcode

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

基本思路：

将树的每一层节点用next串起来。这样每一层也会形成一个单链表。

而每层的链表头，则是，根的左孩子，左孩子，左孩子。

利用双循环，外层循环，沿着根的左孩子，一直向下。

内层循环，负责将下一层的节点串起来。

即，将自己右孩子放到左孩子的next上，而右孩子，则可通过自己的next指针，找到右邻居。

在 leetcode上实际执行时间为25ms。

/**
 * 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 && root->left) {
            TreeLinkNode *p = root;
            while (p) {
                p->left->next = p->right;
                if (p->next)
                    p->right->next = p->next->left;
                
                p = p->next;
            }
            root = root->left;
        }
    }
};

Populating Next Right Pointers in Each Node -- leetcode

原文：http://blog.csdn.net/elton_xiao/article/details/45673841