【second】Populating Next Right Pointers in Each Node II

问题的本质就是：树的level-order遍历

 TreeLinkNode* getNextSibling(TreeLinkNode* cur)
 {
     while(cur)
     {
         if(cur->left)
            return cur->left;
         if(cur->right)
            return cur->right;
         cur = cur->next;
     }
     return NULL;
 }
class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        TreeLinkNode* first,*cur;
        first = root;
        while(first)
        {
            cur = first;
            while(cur)
            {
                if(cur->left&&cur->right)
                {
                    cur->left->next = cur->right;
                    cur->right->next = getNextSibling(cur->next);
                }else if(cur->left&&!cur->right)
                {
                    cur->left->next = getNextSibling(cur->next);
                }else if(!cur->left&&cur->right)
                {
                    cur->right->next = getNextSibling(cur->next);
                }
                cur = cur->next;
            }
            if(first->left)
                first = first->left;
            else if(first->right)
                first = first->right;
            else
                first = getNextSibling(first->next);
        }
        
    }
};

　　

上述解法有大量重复。主要集中在getNextSibling。

    void connect(TreeLinkNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        TreeLinkNode* first,*cur;
        first = root;
        while(first)
        {
            TreeLinkNode* nextFirst = NULL;  //start node of next level
            TreeLinkNode* prev = NULL;　　　　//prev node of cur level
            cur = first;
            while(cur)
            {
                if(!nextFirst)
                    nextFirst = (cur->left?cur->left:cur->right);
                if(cur->left)
                {
                    if(prev)
                        prev->next = cur->left;
                        
                    prev = cur->left;
                }
                
                if(cur->right)
                {
                    if(prev)
                        prev->next = cur->right;
                        
                    prev = cur->right;
                }
                
                cur = cur->next;
            }
            
            first = nextFirst;
        }
        
    }