uva 699(二叉树)

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题解：这道题在建树时要把位置的参数传递以更新每叠叶子的和，题目中有说明叠数不会超过80，就从位置40开始把和存到数组。

#include <stdio.h>#include <string.h>const int N = 85;struct Node {    int val;    Node* left;    Node* right;    Node () {        val = -1;        left = right = NULL;    }};int ans1[N], ans2[N];Node* build(int pos, int a) {    int n;    char c;    Node* node = new Node;    scanf("%d%c", &n, &c);    if(n != -1) {        node -> val = n;        ans1[pos] += n;        node -> left = build(pos - 1, 1);        node -> right = build(pos + 1, 2);    }    else {        if (a == 1)            pos += 1;        else if (a == 2)            pos -= 1;        return NULL;    }    return node;}void Delete(Node* node) {    if (node == NULL)        return;    if (node -> left != NULL)        Delete(node -> left);    if (node -> right != NULL)        Delete(node -> right);    delete node;}int main() {    int cases = 1;    char c;    Node* root;    while (1) {        memset(ans1, 0, sizeof(ans1));        root = build(40, 0);        if (root == NULL)            break;        else {            printf("Case %d:\n", cases++);            int k = 0;            for (int i = 0; i < N; i++)                if (ans1[i] > 0) {                    ans2[k++] = ans1[i];            }            for (int i = 0; i < k - 1; i++)                printf("%d ", ans2[i]);            printf("%d\n\n", ans2[k - 1]);            Delete(root);        }    }    return 0;}

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