【Openjudge】重建二叉树

最新推荐文章于 2019-03-17 08:29:04 发布

原创最新推荐文章于 2019-03-17 08:29:04 发布 · 573 阅读

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本文深入探讨了如何从给定的前序遍历和中序遍历序列重建二叉树的问题，阐述了重建过程的关键步骤与算法实现，帮助读者理解二叉树的构造原理。

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重建

#include<iostream>
#include<stack>
#include<string>

using namespace std;

struct _node{
	char data;
	_node *left, *right;
	_node():data(0), left(NULL), right(NULL){};
};

void build_tree(_node *root, string &preOrders, string &inOrders, int &preOrderpos, int inOrderleft, int inOrderright){
	root->data = preOrders[preOrderpos];
	int inOrdermid = inOrders.find(root->data);
	if (inOrderleft < inOrdermid){
		root->left = new _node;
		preOrderpos ++ ;
		build_tree(root->left, preOrders, inOrders, preOrderpos, inOrderleft, inOrdermid);
	}
	if (inOrdermid < inOrderright - 1){
		root->right = new _node;
		preOrderpos ++;
		build_tree(root->right, preOrders, inOrders, preOrderpos, inOrdermid + 1, inOrderright);
	}
}

int main()
{
	string preOders, inOrders;

	while (cin >> preOders >> inOrders){

		_node* root = new _node;
		int pos = 0;
		build_tree(root, preOders, inOrders, pos, 0, preOders.length());

		//pastOrder travel
		stack<_node*> s_;
		stack<int> tag;
		_node* pointer = root;
		int t;
		while (!s_.empty() || pointer){
			while(pointer){
				s_.push(pointer);
				tag.push(1);
				pointer = pointer->left;
			}
			pointer = s_.top(); s_.pop();
			t = tag.top(); tag.pop();
			if (t == 1 && pointer->right){
				s_.push(pointer);
				tag.push(2);
				pointer = pointer->right;
			}
			else{
				cout << pointer->data;
				pointer = NULL;
			}
		}
		cout << endl;
	}


}