Tree UVA - 548 树的遍历--已知先序中序求权值。

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

方法一：逐个找根节点，直到没有子树为止。

 

#include <iostream>
#include <string>
#include <sstream>
using namespace std; 
const int MAX=10000;
int inf=1e8;
int in[MAX],post[MAX];
int len,minN,minS; 
bool readLine(int *a){
	string s;
	if(!getline(cin,s)) return false;
	stringstream ss(s);
	len=0;
	int n;
	while(ss>>n) a[len++]=n;
	return len>0;
}

int build(int l1,int r1,int l2,int r2,int sum){
	//空树。 
	if(l1>r1) return 0;		
	int root=post[r2];
	sum+=root;
	//叶子节点。 
	if(l1==r1){
		if( minS>sum || (minS==sum && minN>root)){
			minS=sum;
			minN=root;
		}		
		return 0;
	}	
	//在先序遍历中找到根节点。左边为左分支，右边为右分支。 
	int p=l1;
	while(in[p]!=root) p++;
	//左分支个数。 
	int cnt=p-l1;
	build(l1,p-1,l2,l2+cnt-1,sum);
	build(p+1,r1,l2+cnt,r2-1,sum);
	return root;
}

int main(int argc, char** argv) {
	while(readLine(in)){
		minN=inf,minS=inf; 
		readLine(post);
		build(0,len-1,0,len-1,0);
		cout<<minN<<endl;
	}	
	return 0;
}

 方法二：先还原树，再深度遍历

#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
using namespace std;
//因为各个结点的权值各不相同且都是正整数，直接用权值作为结点编号
const int maxv = 10000 + 10;
int in_order[maxv], post_order[maxv], lch[maxv], rch[maxv];
int n;
bool read_list(int* a) {
	string line;
	if(!getline(cin, line)) return false;
	stringstream ss(line);
	n = 0;
	int x;
	while(ss >> x) a[n++] = x;
	return n > 0;
}
//把in_order[L1..R1]和post_order[L2..R2]建成一棵二叉树，返回树根
int build(int L1, int R1, int L2, int R2) {
	if(L1 > R1) return 0; //空树
	int root = post_order[R2];
	int p = L1;
	while(in_order[p] != root) p++;
	int cnt = p-L1; //左子树的结点个数
        //去掉已经选择的节点。
	lch[root] = build(L1, p-1, L2, L2+cnt-1);
	rch[root] = build(p+1, R1, L2+cnt, R2-1);
	return root;
}
int best, best_sum; //目前为止的最优解和对应的权和
void dfs(int u, int sum) {
	sum += u;
	if(!lch[u] && !rch[u]) { //叶子
		if(sum < best_sum || (sum == best_sum && u < best)) { 
		best = u; best_sum= sum; 
		}
	}
	if(lch[u]) dfs(lch[u], sum);
	if(rch[u]) dfs(rch[u], sum);
}
int main() {
	while(read_list(in_order)) {
		read_list(post_order);
		build(0, n-1, 0, n-1);
		best_sum = 1000000000;
		dfs(post_order[n-1], 0);
		cout << best <<endl;
	}
	return 0;
}