Populating Next right pointers on each Node (DFS, not expected)

Given a binary tree

    struct Treelinknode {      treelinknode *left;      Treelinknode *right;      Treelinknode *next;    }

Populate each of the next pointer to the next right node. If There is no next right node, the next pointer should are set to NULL .

Initially, all next pointers is set to NULL .

Note:

• Constant extra space.
• You could assume that it was a perfect binary tree (ie, all leaves was at the same level, and every parent had both children).

For example,
Given the following perfect binary tree,

         1       /        2    3     /\  /     4  5  6  7

After calling your function, the tree is should look like:

         1, NULL       /        2, 3, null     /\  /     4->5->6->7, NULL

It's hard to think of the easy, unexpected.

/** 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) {}}; */classSolution { Public:    voidConnect (Treelinknode *root) {        if(root) {if(root->Left ) Root->left->next = root->Right ; if(root->Right ) Root->right->next = Root->next? Root->next->Left:null; Connect (Root-Left ); Connect (Root-Right ); }    }};

You can directly use the results from the previous layer to DFS.

Reference: http://blog.csdn.net/pickless/article/details/12027997

Populating Next right pointers in each Node (DFS, not expected)