例题6-8 树 UVa548

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1.题目描述：点击打开链接

2.解题思路：本题给出了一颗二叉树的中序遍历和后序遍历，要求找一个叶子，使得它到达根结点的权和最小，如果有多解，那么该叶子自身的权应该尽量小。首先，根据中序遍历和后序遍历建立二叉树，这道题采用数组来存放左右子树的结点值，根为root的左子树结点为lch[root]右子树结点为rch[root]。

那么，如何根据中序遍历，后序遍历来建树呢？方法是根据后序遍历找到根，然后在中序遍历中找到树根，从而找出了左右子树的结点列表，然后递归构造左右子树即可。最后利用先序遍历来求解最优的叶子结点。

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define N 10000+10int in_order[N], post_order[N], lch[N], rch[N];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;}int build(int L1, int R1, int L2, int R2)//建树{if (L1 > R1)return 0;int root = post_order[R2];int p = L1;while (in_order[p] != root)p++;int cnt = p - L1;lch[root] = build(L1, p - 1, L2, L2 + cnt - 1);//递归建左子树rch[root] = build(p + 1, R1, L2 + cnt, R2 - 1);//递归建右子树return root;}int best, best_sum;void dfs(int u, int sum)//先序遍历找最优叶子{sum += u;if (!lch[u] && !rch[u]){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(){//freopen("t.txt", "r", stdin);while (read_list(in_order)){read_list(post_order);build(0, n - 1, 0, n - 1);best_sum = 100000000;dfs(post_order[n - 1], 0);cout << best << endl;}return 0;}

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