剑指offer-面试题6：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，请重建该二叉树。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。

解：算法的关键是找出根节点的左子树和右子树。如果算法能求出根节点和左右子树，剩下的递归调用这个算法就可以了。前序遍历的第一个值就是根节点，而中序遍历中左子树就是根节点前面的部分，右子树就是根节点右边的部分，问题得解。

下面是代码及测试部分。

#include <iostream>
using namespace std;
struct BinaryTreeNode
{
	int m_nValue;
	BinaryTreeNode* m_pLeft;
	BinaryTreeNode* m_pRight;
};

BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder)
{
	int rootValue = startPreorder[0];
	BinaryTreeNode* root = new BinaryTreeNode();
	root->m_nValue = rootValue;
	root->m_pLeft = root->m_pRight = NULL;
	
	if(startPreorder == endPreorder)
	{
		if(startInorder == endInorder && *startPreorder == *startInorder)
			return root;
		else
			cout << "Invalid input." << endl;
	}
	
	int* rootInorder = startInorder;
	while(rootInorder <= endInorder && *rootInorder != rootValue)
		++rootInorder;
	if(rootInorder == endInorder && *rootInorder != rootValue)
		cout << "Invalid input." << endl;

	int leftLength = rootInorder - startInorder;
	int* leftPreorderEnd = startPreorder + leftLength;
	if(leftLength > 0)
		root->m_pLeft = ConstructCore(startPreorder+1, leftPreorderEnd, startInorder, rootInorder-1);
	if(leftLength < endPreorder - startPreorder)
		root->m_pRight = ConstructCore(leftPreorderEnd+1, endPreorder, rootInorder+1, endInorder);
	return root;
}

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);
}

void PreorderTraversal(BinaryTreeNode* T)
{
	if(T)
	{
 		cout << T->m_nValue << " ";
		PreorderTraversal(T->m_pLeft);
		PreorderTraversal(T->m_pRight);
	}
	else
		return;
}

void InorderTraversal(BinaryTreeNode* T)
{
	if(T)
	{
		InorderTraversal(T->m_pLeft);
		cout << T->m_nValue << " ";
		InorderTraversal(T->m_pRight);
	}
	else
		return;
}


int main() {
	// your code goes here
	int preOrder[] = {1,2,4,7,3,5,6,8};
	int inOrder[] = {4,7,2,1,5,3,8,6};
	BinaryTreeNode* tree = Construct(preOrder, inOrder, 8);
	PreorderTraversal(tree);
	cout << endl;
	InorderTraversal(tree);
	return 0;
}

延伸：如果知道中序遍历和后序遍历呢？因为根节点为后续遍历的最后一个数，所以依然可以从中序遍历中划分出根节点的左右子树。如果知道前序遍历和后续遍历呢，能不能重建二叉树？我认为（仅仅是个人认为）是不能的，因为这时找不到左右子树的边界。

下面是知道中序遍历和后续遍历重建二叉树的代码：

BinaryTreeNode* ConstructCore(int* startInorder, int* endInorder, int* startPostorder, int* endPostorder)
{
	int rootValue = *endPostorder;
	BinaryTreeNode* root = new BinaryTreeNode();
	root->m_nValue = rootValue;
	root->m_pLeft = root->m_pRight = NULL;
	
	if(startInorder == endInorder)
	{
		if(startPostorder == endPostorder && *startInorder == *startPostorder)
			return root;
		else
			cout << "Invalid input." << endl;
	}
	
	int* rootInorder = startInorder;
	while(rootInorder <= endInorder && *rootInorder != rootValue)
		++rootInorder;
	if(rootInorder == endInorder && *rootInorder != rootValue)
		cout << "Invalid input." << endl;

	int leftLength = rootInorder - startInorder;
	int* leftPostorderEnd = startPostorder + leftLength - 1;
	if(leftLength > 0)
		root->m_pLeft = ConstructCore(startInorder, rootInorder-1, startPostorder, leftPostorderEnd);
	if(leftLength < endInorder - startInorder)
		root->m_pRight = ConstructCore(rootInorder+1, endInorder, leftPostorderEnd+1, endPostorder-1);
	return root;
}

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