剑指offer(15)-二叉搜索树与双向链表

题目：

输入一棵二叉搜索树，将该二叉搜索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，只能调整树中结点指针的指向。

二叉树结点的定义如下：

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

class Solution {
public:
TreeNode* Convert(TreeNode* pRootOfTree)
{
TreeNode *pLastNodeInList = NULL;
ConvertNode(pRootOfTree,&pLastNodeInList);
TreeNode *pHeadOfList = pLastNodeInList;
while (pHeadOfList != NULL && pHeadOfList->left != NULL)
pHeadOfList = pHeadOfList->left;

return pHeadOfList;

}

void ConvertNode(TreeNode* pNode,TreeNode **pLastNodeInList)
{
if (pNode == NULL)
return;
TreeNode *pCurrent = pNode;

if (pCurrent->left != NULL)
ConvertNode(pCurrent->left,pLastNodeInList);

pCurrent->left = *pLastNodeInList;
if (*pLastNodeInList != NULL)
(*pLastNodeInList)->right = pCurrent;

*pLastNodeInList = pCurrent;

if (pCurrent->right != NULL)
ConvertNode(pCurrent->right,pLastNodeInList);
}
};

另外Crack Coding interview上有一道相关的BST题目，在这里贴一下:

对于一个元素各不相同且按升序排列的有序序列，请编写一个算法，创建一棵高度最小的二叉查找树。

给定一个有序序列int[] vals,请返回创建的二叉查找树的高度。

import java.util.*;
/*
class TreeNode
{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int data)
{
left = null;
right = null;
val = data;
}
}
*/

public class MinimalBST {

TreeNode buildBST(int [] vals,int left,int right)
{

if (left > right)
return null;
int middle = left + (right - left) / 2;
TreeNode root = new TreeNode(vals[middle]);
root.left = buildBST(vals,left,middle - 1);
root.right = buildBST(vals,middle + 1,right);
return root;
}

int highBST(TreeNode root)
{
if (root == null)
return 0;
int left = highBST(root.left);
int right = highBST(root.right);
if (left > right)
return left + 1;
else
return right + 1;
}

int buildMinimalBST(int [] vals)
{
int length = vals.length;
if (length <= 0)
return 0;
TreeNode root = buildBST(vals,0,length - 1);
return highBST(root);
}
}