Leetcode 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 toNULL.

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

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写代码要考虑缜密，数据的初始化状态，数据为空的处理，循环的判断条件都考虑清楚再写，而不是想到什么再写什么。无论做什么事都需要提前规划，考虑各种可能出现的情况，要心思缜密。本题实际是就是层次遍历，但一般层次遍历需要借助栈等数据结构，不符合要求。但每个节点多了一个next指针，可以使用该指针来替代原先的栈等数据结构。

代码如下：

/** * 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) {       TreeLinkNode* cur=root;        while(cur)        {            TreeLinkNode* temp = cur;            while(temp)            {                if(temp->left != NULL)                    temp->left->next =temp->right;                if(temp->right != NULL && temp->next != NULL)                    temp->right->next = temp->next->left;                temp = temp->next;            }            cur = cur->left;                    }    }};

也可以使用递归大法，代码如下：

/** * 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) {       if(!root)           return;                if(root->left)            root->left->next = root->right;        if(root->right)        {            if(root->next)                root->right->next = root->next->left;            else                root->right->next = NULL;        }        connect(root->left);        connect(root->right);    }};

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