106 lines
2.5 KiB
C
106 lines
2.5 KiB
C
#include <stdio.h>
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#include <stdlib.h>
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int n;
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int m;
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// 二叉排序树，根节点为 1，左子树为 2 * i，右子树为 2 * i + 1
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int tree[1000];
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void makeTree(int val) {
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int r = 1;
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while (1) {
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if (val < tree[r]) {
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if (tree[2 * r] == 0) {
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// printf("r: %d, val: %d\n", 2 * r, val);
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tree[2 * r] = val;
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break;
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} else {
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r = 2 * r;
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}
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} else {
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if (tree[2 * r + 1] == 0) {
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// printf("r: %d, val: %d\n", 2 * r + 1, val);
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tree[2 * r + 1] = val;
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break;
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} else {
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r = 2 * r + 1;
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}
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}
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}
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}
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// 中序遍历
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void inOrder(int root) {
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if (tree[2 * root] != 0)
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inOrder(2 * root);
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printf("%d ", tree[root]);
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if (tree[2 * root + 1] != 0)
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inOrder(2 * root + 1);
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}
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// 二叉排序树T查找成功的平均查找长度
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double ASL(int root, int level) {
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if (tree[root] == 0)
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return 0;
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return level + ASL(2 * root, level + 1) + ASL(2 * root + 1, level + 1);
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}
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// 找到二叉树中值为val的节点并删除
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void delNode(int root, int x) {
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if (tree[root] == 0)
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return;
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if (tree[root] == x) {
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if (tree[2 * root] == 0 && tree[2 * root + 1] == 0) {
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tree[root] = 0;
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} else if (tree[2 * root] != 0 && tree[2 * root + 1] == 0) {
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tree[root] = tree[2 * root];
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tree[2 * root] = 0;
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} else if (tree[2 * root] == 0 && tree[2 * root + 1] != 0) {
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tree[root] = tree[2 * root + 1];
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tree[2 * root + 1] = 0;
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} else {
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int r = 2 * root + 1;
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while (tree[2 * r] != 0) {
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r = 2 * r;
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}
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tree[root] = tree[r];
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tree[r] = 0;
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}
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} else if (tree[root] > x) {
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delNode(2 * root, x);
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} else {
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delNode(2 * root + 1, x);
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}
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}
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void pt() {
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printf("inOrder:\n");
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inOrder(1);
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printf("\n");
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printf("ASL: %lf\n", ASL(1, 1) / n);
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printf("\n");
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}
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int main() {
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freopen("data.in", "r", stdin);
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scanf("%d", &n);
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for (int i = 1; i <= n; i++) {
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int val;
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scanf("%d", &val);
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if (i == 1) {
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tree[1] = val;
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} else {
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makeTree(val);
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}
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}
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pt();
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scanf("%d", &m);
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for (int i = 0; i < m; i++) {
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int x;
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scanf("%d", &x);
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delNode(1, x);
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pt();
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}
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return 0;
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} |