leetcode: Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1       /  \      2    3     / \    \    4   5    7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \    \    4-> 5 -> 7 -> NULL

/** * 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 == NULL)            return;        if( root->left && root->right){            root->left->next = root->right;            root->right->next = nextNode( root->next);            connect( root->right);//这里的顺序很关键，要先递归右子树再递归左子树，不然当左子树构建时右子树还没中next指针还没构建            connect( root->left);        }        else if( root->left){            root->left->next = nextNode( root->next);            connect( root->left);        }        else if( root->right){            root->right->next = nextNode( root->next);            connect( root->right);        }        else            return;               }    TreeLinkNode *nextNode( TreeLinkNode * root){        if( root == NULL)            return NULL;        if( root->left)            return root->left;        else if( root->right)            return root->right;        else            return nextNode( root->next);    }};