Populating Next Right Pointers in Each Node II 题解

题目来源：Populating Next Right Pointers in Each Node II

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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

解题思路：

跟 Populating Next Right Pointers in Each Node思路一致，值得注意的是

• 在填充下一层next时，当前层的next要找完。//Attention 0

• 往下走时，可能前面的节点没有孩纸节点，也要找完同一层的节点是否存在孩纸节点。//Attention 1

void connect(TreeLinkNode *root) 
{
   while(root)
   {
       auto down = root->left;
       if(NULL == down) down = root->right;
       if(NULL == down) down = root->next;////Attention 1: no child of root, but maybe the next(or next') node has children, although the same level
       while(root)//go right
       {
           auto next = root->next; //go right to find next's children
           if(root->left)
           {
               if(root->right)
                   root->left->next = root->right;
               else{
                   while (next && (next->left == NULL && next->right == NULL)) //Attention 0
                       next = next->next;
                   if(next != NULL)
                       root->left->next = next->left == NULL ? next->right : next->left;
               }
           }
           if(root->right)
           {
               while (next && (next->left == NULL && next->right == NULL))
                   next = next->next;
               if(next != NULL)
                   root->right->next = next->left == NULL ? next->right : next->left;
           }
           root = next;
       }
       root = down;//go down
   }    
}

上面代码确实比较丑陋～ 其实着眼于同一层之前通过pre来记录同一层的前一个节点的话，代码就好看得多。 leetcode-cpp

void connect(TreeLinkNode *root)
{
    while(root)
    {
        TreeLinkNode * next = NULL;//next level
        TreeLinkNode * pre = NULL;
        while(root)//go right
        {
            if(next == NULL) next = root->left ? root->left : root->right;
            if(root->left)
            {
                if(pre) pre->next = root->left;
                pre = root->left;
            }
            if(root->right)
            {
                if(pre) pre->next = root->right;
                pre = root->right;
            }
            root = root->next;
        }
        root = next;//goto next level
    }
}