二叉查找树转变为有序双向链表

题目：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。要求不能创建任何新的结点，只调整指针的指向。

　　比如将二元查找树
                                         10
                                        /    \
                                      6       14
                                    /  \     /　 \
                                 4     8  12   16
转换成双向链表

4=6=8=10=12=14=16。

一道微软的面试题。

二叉查找树的每个节点都有两个指针，双向链表的节点也是两个指针，不创建新节点只调整指针是可以完成转变的。

对于二叉查找树的每个节点Node，它的左子树中所有的关键字都小于Node的关键字，而右子树中的所有关键字都大于Node的关键字。

采用递归的思维可以很方便的完成转变，将节点的左右子树分别转变成有序链表，然后将节点的指针分别指向这两个链表。

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import random class Node(object):     def __init__(self,key):         self.key=key         self.left=None         self.right=None class BSTree(object):     def __init__(self):         self.root=None     def put(self,key):         if not self.root:             self.root=Node(key)         else:             self.root=self._put(self.root,key)     def _put(self,node,key):         if node is None:             node=Node(key)         elif key<node.key:             node.left=self._put(node.left,key)         elif key>node.key:             node.right=self._put(node.right,key)         return node      def convert(self):         if self.root:             return self._convert(self.root)     def _convert(self,node,asright=True):         if not node:             return None         else:             left=self._convert(node.left,False)             if left:                 left.right=node                 node.left=left             right=self._convert(node.right)             if right:                 right.left=node                 node.right=right             cur=node             if asright:                 while cur.left:                     cur=cur.left             if not asright:                 while cur.right:                     cur=cur.right             return cur                                 if __name__=='__main__':     t=BSTree()       for i in range(10):         t.put(random.randint(0,100))        cur=t.convert()     if cur:         print cur.key         while cur.right:             cur=cur.right             print cur.key         while cur.left:             cur=cur.left             print cur.key

　　另一种思路是采用中序遍历的方法。二叉查找树中序遍历的话就是从小到大排列。将遍历过的节点转为有序链表，然后将下一个要遍历的节点加到链表结尾。

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import random class Node(object):     def __init__(self,key):         self.key=key         self.left=None         self.right=None class Sorted_LinkedList(object):     def __init__(self):         self.head=None     def travel(self):         node=self.head         while node:             print node.key             node=node.right class BSTree(object):     def __init__(self):         self.root=None         self.list=Sorted_LinkedList()         self.curNode=None     def put(self,key):         if not self.root:             self.root=Node(key)         else:             self.root=self._put(self.root,key)     def _put(self,node,key):         if node is None:             node=Node(key)         elif key<node.key:             node.left=self._put(node.left,key)         elif key>node.key:             node.right=self._put(node.right,key)         return node      def convert(self):         self._travel(self.root)         return self.list     def _travel(self,node):         if node:              self._travel(node.left)              if self.curNode:                  self.curNode.right=node                  node.left=self.curNode              else:                  self.list.head=node              self.curNode=node              self._travel(node.right)                           if __name__=='__main__':     t=BSTree()       for i in range(100):         t.put(random.randint(0,100))     l=t.convert()     l.travel()