UVA - 548 Tree

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题目大意：给一棵树的中序遍历和后序遍历，找一叶子到根的值最小的叶子，若有多解，输出叶子本身最小的那个叶子。

解题思路：根据中序遍历和后序遍历建树。后续的最后一个数是根节点，该数在中序中的前一个数是该数的左孩子，cnt 是左子树节点个数。dfs 求解。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<ctype.h>using namespace std;int middle[10010], after[10010], lch[10010], rch[10010];int n, maxans, ans;int build(int l1, int r1, int l2, int r2) {    if (l1 > r1) return 0;    int root = after[r2];    int tag = l1;    while (middle[tag] != root) tag++;    int cnt = tag-l1;    lch[root] = build(l1, tag-1, l2, l2+cnt-1);    rch[root] = build(tag+1, r1, l2+cnt, r2-1);    return root;}void dfs(int leaf, int sum) {    sum += leaf;    if (!lch[leaf] && !rch[leaf]) {        if (sum < maxans || (sum == maxans && leaf < ans)) {            ans = leaf;            maxans = sum;        }    }        if (lch[leaf]) dfs(lch[leaf], sum);        if (rch[leaf]) dfs(rch[leaf], sum);}int main() {    char c;    while (scanf("%d%c", &middle[0],&c) != EOF) {    n = 1;    while (c != '\n') scanf("%d%c", &middle[n++], &c);    for (int i = 0; i < n; i++)        scanf("%d", &after[i]);    build(0, n-1, 0, n-1);    maxans = 999999999;    dfs(after[n-1], 0);    printf("%d\n", ans);    }return 0;}

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