Flatten Binary Tree to Linked List

题目：

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1
        / \
       2   5
      / \   \
     3   4   6

 

The flattened tree should look like:

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

其实就是把二叉树按前序遍历顺序以右指针为next搞成一个单链表，想了半天前驱、后继貌似没折，后来想把左子树先平掉，再把左子树合到右子树顶端之上，再平掉原先右子树的那个结点为根的树即可。代码如下：

    void flatten(TreeNode *root) {
        if(root==NULL) return;
        TreeNode* t,*right=root->right;
        flatten(root->left);//平掉左子树
        if(root->left){//把左子树插入到右子树与根之间
            t = root->left;
            while(t&&t->right) t=t->right;
            right = t->right = root->right;
            root->right=root->left;
            root->left=NULL;
        }
        flatten(right);//平掉右子树
    }