uva 548 tree（中序后序还原树）

最新推荐文章于 2018-09-20 01:31:31 发布

JeraKrs

最新推荐文章于 2018-09-20 01:31:31 发布

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分类专栏： UVA 数据结构-二叉树训练指南-第三章

本文链接：https://blog.youkuaiyun.com/keshuai19940722/article/details/9341605

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训练指南-第三章

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该博客介绍了如何通过中序和后序遍历重建二叉树，并找到从根到叶子节点的最小路径和，输出对应的叶子节点值。解题策略涉及二叉树的构建与遍历。

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Tree

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values of nodes along that path.

Input

The input file will contain a description of the binary tree given as the inorder and postorder traversal sequences of that tree. Your program will read two line (until end of file) from the input file. The first line will contain the sequence of values associated with an inorder traversal of the tree and the second line will contain the sequence of values associated with a postorder traversal of the tree. All values will be different, greater than zero and less than 10000. You may assume that no binary tree will have more than 10000 nodes or less than 1 node.

Output

For each tree description you should output the value of the leaf node of a path of least value. In the case of multiple paths of least value you should pick the one with the least value on the terminal node.

Sample Input

3 2 1 4 5 7 6
3 1 2 5 6 7 4
7 8 11 3 5 16 12 18
8 3 11 7 16 18 12 5
255
255

Sample Output

1
3
255

题目大意：给出中序和后序，求出由根到叶子的路径中，和最小的，输出叶子的值（如果有相同的，输出叶子值较小的）

解题思路：由中序和后序还原树，由根向下层层遍历，比较最后叶子的值，输出该叶子原先的值。

#include<stdio.h>
#include<string.h>

#define N 100005
struct node{
	node(int a)
	{
		sum=0;
		num=a;
		lift=NULL;
		right=NULL;
	}
	int sum;
	int num;
	node *lift;
	node *right;
};

int p[N];
int q[N];
int min,cnt;
int move;

int find(int k)
{	int i;
	for(i = 0;;i++)
		if(p[i]==k)
			return i;
}

node* grow(int a,int b);
void print(node *f);

int main()
{
	char c;
	int i;

	//Init.
	memset(p,0,sizeof(p));
	memset(q,0,sizeof(q));
	

	while(scanf("%d%c",&p[0],&c)==2)
	{
		min=1000000;
		//Read.
		for(i = 1;c != '\n'; i++)
			scanf("%d%c",&p[i],&c);
		
		int k;
		c='\0';
		for(k = 0;c != '\n'; k++)
			scanf("%d%c",&q[k],&c);

		move=k-1;
		int t=find(q[move]);
		node *head;
		head=new node(q[move--]);
		head->right=grow(t,k);
		head->lift=grow(-1,t);

		head->sum=head->num;
		print(head);
		printf("%d\n",cnt);
	}
	return 0;}

node *grow(int a,int b)
{
	if(a == b-1)	return NULL;
	else
	{
		int t=find(q[move]);
		node *f;
		f=new node(q[move--]);
		f->right=grow(t,b);
		f->lift=grow(a,t);
	
		return f;
	}
}
void print(node *f)
{
	if(f->lift==NULL&&f->right==NULL)
	{	if(min > f->sum)
		{	min=f->sum;
			cnt=f->num;
		}
		else if(min == f->sum&& cnt > f->num)
			cnt=f->num;

	}
	else
	{
		if(f->lift!=NULL)
		{
			f->lift->sum=f->lift->num+f->sum;
			print(f->lift);
		}

		if(f->right!=NULL)
		{	f->right->sum=f->right->num+f->sum;
			print(f->right);
		}
	}
	delete f;
}