微软面试100题---将 二叉搜索树 转化成 有序的双向链表

Description:

输入一棵二元查找树,将该二元查找树转换成一个排序的双向链表。

二叉搜索树和双向链表的定义与实现参考: http://blog.csdn.net/shoulinjun/article/details/18449445

                                                                 http://blog.csdn.net/shoulinjun/article/details/18324217

要求不能创建任何新的结点,只调整指针的指向。
         10
       /       \

     6        14

  /      \    /      \

4       8 12    16

转换成双向链表

4=6=8=10=12=14=16

数据结构如下：

struct TreeNode{
  int val;
  TreeNode *left;
  TreeNode *right;
  TreeNode(int x)
    :val(x), left(NULL), right(NULL){}
};

递归：

TreeNode* Convert(TreeNode *root)
{ /* exit */
  if(root == NULL) return root;

  if(root->right){
    root->right = Convert(root->right);
  }

  if(root->left){
    TreeNode *leftHead = Convert(root->left);
    TreeNode *leftRear = leftHead;
    for(; leftRear->right; leftRear = leftRear->right);

    /* merge */
    leftRear->right = root;
    root->left = leftRear;
    root = leftHead;
  }

  return root;
}

非递归，采用 stack, 来模拟递归。思想的核心其实就是 中序遍历的非递归版本（栈），参考 http://blog.csdn.net/shoulinjun/article/details/19089155

TreeNode *ConvertNonRecur(TreeNode *root)
{
  TreeNode *head(NULL), *rear(NULL);
  stack<TreeNode*> s;

  while(root || !s.empty())
  {
    while(root){
	  s.push(root);
	  root = root->left;
	}
	root = s.top(); s.pop();
	TreeNode *next = root->right;

	//insert root into double-linked-list
	if(rear){
	  root->right = rear->right;
	  rear->right = root;
	  root->left = rear;
	  rear = rear->right;
	}
	else{
	  head = rear = root;
	  head->right = head->left = NULL; 
	}
	root = next;
  }

  return head;
}