[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

思路：与Populating Next Right Pointers in Each Node类似，采用递归，只是需要处理一些特殊情况

代码：

/** * 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 *temp;        int flag=0;        if(root == NULL) return;          if(root->left != NULL){            temp=root;            flag=0;            while(temp!=NULL){                if(temp->left!=NULL && temp->left!=root->left){                     root->left->next=temp->left;                    flag=1;                    break;                }                if(temp->right!=NULL){                     root->left->next=temp->right;                    flag=1;                    break;                }                temp=temp->next;            }            if(flag==0) root->left->next=NULL;        }          if(root->right !=NULL){            flag=0;            temp=root->next;            while(temp!=NULL){                if(temp->left!=NULL){                     root->right->next=temp->left;                    flag=1;                    break;                }                if(temp->right!=NULL){                     root->right->next=temp->right;                    flag=1;                    break;                }                temp=temp->next;            }            if(flag==0) root->right->next=NULL;        }       connect(root->right);         connect(root->left);      }};