二叉树的重建

1.根据二叉树的中序和后续遍历重建二叉树

思路：假设递归过程中，某一步的后序序列区间为[postL,postR]，中序序列区间为[inL,inR]；

1. 根据后序遍历的特点可知，postR位置为根结点；

2. 从中序序列中，寻找出root的位置k，k左边的均为左子树，右边的均为右子树；

3. 将左子树区间[postL,k-1]作为新的后序和中序序列，进行递归。

代码如下:

#include 
#include 
#include 
using namespace std;

struct BTNode {
	BTNode(char data):_data(data),_pLeft(nullptr),_pRight(nullptr){}
	char _data;
	BTNode* _pLeft;
	BTNode* _pRight;
};

void PreOrder(BTNode* pRoot) {
	if (pRoot == nullptr)
		return;
	cout << pRoot->_data << " ";
	PreOrder(pRoot->_pLeft);
	PreOrder(pRoot->_pRight);
}

void PostOrder(BTNode* pRoot) {
	if (pRoot == nullptr)
		return;
	PostOrder(pRoot->_pLeft);
	PostOrder(pRoot->_pRight);
	cout << pRoot->_data << " ";
}

BTNode* ReBuildBTTree(string& Post, string& In, int PostL, int PostR, int InL, int InR) {
	if (PostL > PostR)
		return nullptr;
	BTNode* pRoot = new BTNode(Post[PostR]);
	int k = 0;
	while (In[k] != pRoot->_data) {
		++k;
	}
	int numLeft = k - InL;
	pRoot->_pLeft = ReBuildBTTree(Post, In, PostL, PostL + numLeft - 1, InL, k - 1);
	pRoot->_pRight = ReBuildBTTree(Post, In, PostL + numLeft, PostR - 1, k + 1, InR);
	return pRoot;
}
int main() {
	string In = "dgbaechf";
	string Post = "gbdehfca";
	string Pre = "adbgcefh";
	BTNode* pNewRoot = ReBuildBTTree(Post, In, 0, 7, 0, 7);
	PreOrder(pNewRoot);
	cout << endl;
	system("pause");
	return 0;
}

输出结果:
a d b g c e f h

2.根据二叉树的中序和前序遍历重建二叉树

思路：假设递归过程中，某一步的后序序列区间为[preL,preR]，中序序列区间为[inL,inR]；

1. 根据前序遍历的特点可知，postL位置为根结点；

2. 从中序序列中，寻找出root的位置k，k左边的均为左子树，右边的均为右子树；

3. 将左子树区间[preL+1,preL+numLeft]作为新的后序和中序序列，进行递归。

代码如下:

#include 
#include 
#include 
using namespace std;

struct BTNode {
	BTNode(char data):_data(data),_pLeft(nullptr),_pRight(nullptr){}
	char _data;
	BTNode* _pLeft;
	BTNode* _pRight;
};
void PostOrder(BTNode* pRoot) {
	if (pRoot == nullptr)
		return;
	PostOrder(pRoot->_pLeft);
	PostOrder(pRoot->_pRight);
	cout << pRoot->_data << " ";
}
BTNode* Create(string& Pre, string& In, int preL, int preR, int InL, int InR) {
	if (preL > preR)
		return nullptr;
	BTNode* pRoot = new BTNode(Pre[preL]);
	int k = 0;
	while (In[k] != pRoot->_data) {
		++k;
	}
	int numLeft = k - InL;
	pRoot->_pLeft = Create(Pre, In, preL + 1, preL + numLeft, InL, k - 1);
	pRoot->_pRight = Create(Pre, In, preL + numLeft + 1, preR, k + 1, InR);
	return pRoot;
}

int main() {
	string In = "dgbaechf";
	string Post = "gbdehfca";
	string Pre = "adbgcefh";
	BTNode* pNew2 = Create(Pre, In, 0, 7, 0, 7);
	PostOrder(pNew2);
	cout << endl;
	system("pause");
	return 0;
}

输出结果:
g b d e h f c a