二叉树,中序遍历+后序遍历输出前序遍历

引用:

https://blog.csdn.net/m0_37698652/article/details/79218014

#include <iostream>
#include <string>

using namespace std;
struct TreeNode
{
	struct TreeNode* left;
	struct TreeNode* right;
	char  elem;
};


TreeNode* BinaryTreeFromOrderings(const char*inorder, const char*aftorder, int length)
{
	if (length == 0)
	{
		return NULL;
	}
	TreeNode* node = new TreeNode;
	node->elem = *(aftorder + length - 1);//后序遍历的最后一个结点为根结点
	cout << node->elem;//输出当前结点
	int rootIndex = 0;
	for (; rootIndex < length; rootIndex++)//确定根结点位置
	{
		if (inorder[rootIndex] == *(aftorder + length - 1))//中序遍历中,根结点左侧为左子树,右侧为右子树
			break;
	}
	//递归
	node->left = BinaryTreeFromOrderings(inorder, aftorder, rootIndex);
	node->right = BinaryTreeFromOrderings(inorder + rootIndex + 1, aftorder + rootIndex, length - (rootIndex + 1));

	return node;
}


int main()
{
	int n;
	cout << "请输入二叉树结点数量" << endl;
	while (cin >> n) {
		string ins, afs;
		cout << "请输入中序遍历结果(字符间无间隔)" << endl;
		cin >> ins;
		cout << "请输入后序遍历结果(字符间无间隔)" << endl;
		cin >> afs;
		const char*in = ins.data();
		const char*af = afs.data();
		
		cout << "该二叉树的前序遍历结果为" << endl;
		BinaryTreeFromOrderings(in, af, n);
		cout << endl << endl;
		cout << "请输入二叉树结点数量" << endl;
	}
	return 0;
}

猜你喜欢

转载自blog.csdn.net/Perce_Issac/article/details/85003460
今日推荐