剑指offer-面试题6

题目： 输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。

解题思路：递归实现

/**********************
* 面试题6 ： 重建二叉树
***********************/

struct BinaryTreeNode
{
	int								Value;
	BinaryTreeNode*		pLeft;
	BinaryTreeNode*		pRight; 	
};


BinaryTreeNode* Construct(int* preorder, int* inorder, int length)
{
	if(preorder == NULL || inorder == NULL || length <= 0)
		return NULL;
		
	return ConstructCore(preorder, preorder + length - 1,
												inorder, inorder + length - 1);	
}

BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder,
															int* startInorder, int* endInorder)
{
	//前序遍历序列的第一个数字是根节点的值
	int rootValue = startPreorder[0];
	BinaryTreeNode* root = new BinaryTreeNode();
	root->Value = rootValue;
	root->pLeft = root->pRight = NULL;
	
	if(startPreorder == endPreorder)
	{
		if(startInorder = endInorder
			 && *startPreorder == *startInorder)
			 return root;
		else
			throw exception("Invalid input");	
	}	
	
	// 在中序遍历序列中找到根节点的值
	int* rootInorder = startInorder;
	while(rootInorder <= endInorder && *rootInorder != rootValue)
	{
		++rootInorder;	
	}
	
	if(rootInorder == endInorder && *rootInorder != rootValue)
	{
		throw exception("Invalid input");	
	}
	
	int leftLength = rootInorder - startInorder;
	int* leftPreorderEnd  = startPreorder +leftLength;
	if(leftLength > 0)
	{
		//构建左子树
		root->pLeft = ConstructCore(startPreoder + 1, 
																leftPreorderEnd, startInorder, rootInorder -1);	
	}
	if(leftLength < endPreorder - startPreorder)
	{
		//构建右子树
		root->pRight = ConstructCore(leftPreorderEnd + 1，
																	endPreorder, rootInorder + 1, endInorder);	
	}
	
	return root;
}