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

Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

分析

将二叉树转化为前序遍历的链表。使用递归，先处理左子树，然后找到左子树形成的链表的尾端，链接到右子树上，再处理右子树即可。C代码如下已通过。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* preOrder(struct TreeNode* root)
{
    if(root==NULL||(root->left==NULL&&root->right==NULL))
        return root;
    struct TreeNode *l=root->left,*r=root->right,*temp;
    root->left=NULL;
    if(l!=NULL)
    {
        printf("%d ",l->val);
        temp=preOrder(l);
        root->right=temp;
        while(temp->right!=NULL)
            temp=temp->right;
    }
    else
    {
        temp=root;
    }
    temp->right=preOrder(r);
    temp->left=NULL;
    return root;
}
void flatten(struct TreeNode* root) {
    preOrder(root);
}