重建二叉树

/*****************************************************************
题目：输入某二叉树的前序遍历和中序遍历的结果，请重新建出该二叉树。
假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序
序列{1,2,4,7,3,5,6,8}和中序序列{4,7,2,1,5,3,8,6}，则重建出如图所示的
二叉树。二叉树的定义如下：

	1
        /        2   3
      /   / \ 
     4   5   6
      \     /
       7   8
******************************************************************/

#include<stdio.h>
#include<iostream>
struct BinaryTreeNode
{
	int m_nKey;
	BinaryTreeNode* m_pLeft;
	BinaryTreeNode* m_pRight;
};

BinaryTreeNode* createBinaryTreeNode(int value)
{
	BinaryTreeNode* pNode = new BinaryTreeNode();
	pNode->m_nKey = value;
	pNode->m_pLeft = NULL;
	pNode->m_pRight = NULL;
	return pNode;
}

void connectBinaryTreeNode(BinaryTreeNode* pParent,BinaryTreeNode* pLeft,BinaryTreeNode* pRight)
{
	if(pParent)
	{
		pParent->m_pLeft = pLeft;
		pParent->m_pRight = pRight;
	}
}

void destroyBinaryTree(BinaryTreeNode* pRoot)
{
	if(pRoot)
	{
		BinaryTreeNode* pLeft = pRoot->m_pLeft;
		BinaryTreeNode* pRight = pRoot->m_pRight;
		delete pRoot;
		destroyBinaryTree(pLeft);
		destroyBinaryTree(pRight);
	}
}

BinaryTreeNode* concreteBinaryTree(int* preOrder,int* midOrder,int length)
{
	if(preOrder && midOrder && length != 0)
	{
		int i = 0;
		for(i=0; i<length; i++)
		{
			if(preOrder[0] == midOrder[i])
				break;
		}

		if(i>=length)	//如果在中序排列队列中找不到根节点，抛出异常
			throw std::exception("Invalid input");

		BinaryTreeNode* pNode = createBinaryTreeNode(preOrder[0]);
		if(i > 0)
			pNode->m_pLeft = concreteBinaryTree(preOrder+1,midOrder,i);
		if(length-1-i > 0)
			pNode->m_pRight = concreteBinaryTree(preOrder+i+1,midOrder+i+1,length-1-i);	

		return pNode;
	}
	else
		return NULL;

}
void pre_Order(BinaryTreeNode* pRoot)
{
	if(pRoot)
	{
		printf("%d\t",pRoot->m_nKey);
		pre_Order(pRoot->m_pLeft);
		pre_Order(pRoot->m_pRight);		
	}
}
void mid_Order(BinaryTreeNode* pRoot)
{
	if(pRoot)
	{
		mid_Order(pRoot->m_pLeft);
		printf("%d\t",pRoot->m_nKey);
		mid_Order(pRoot->m_pRight);
	}
}

void test()
{
	const int length = 8;
	int preOrder[8] = {1,2,4,7,3,5,6,8};
	int midOrder[8] = {4,7,2,1,5,3,8,6};

	BinaryTreeNode* pNode1 = concreteBinaryTree(preOrder,midOrder,length);
		
	pre_Order(pNode1);
	printf("\n");
	mid_Order(pNode1);
}
int main()
{
	test();
	return 0;
}

重建二叉树,布布扣,bubuko.com

重建二叉树

原文：http://blog.csdn.net/walkerkalr/article/details/20300453