HDU 5444 Elven Postman

来源：互联网发布：大数据产业链包括哪些编辑：程序博客网时间：2024/05/22 17:04

2015长春Online:http://acm.hdu.edu.cn/showproblem.php?pid=5444

建一个二叉树，插入多个点，每次从根节点开始插入，如果当前节点为空则插入当前节点，否则如果小于当前节点的值，则插入右节点，否则插入左节点。

遍历的时候也如此，若找左儿子则输出W，找右儿子输出E，根节点输出空行。

#include <stdio.h>struct tree{tree *lchild;tree *rchild;int key;//tree(){}tree(int x){key = x;lchild = rchild = NULL;}};int q[1005];void build(tree *root, int key){if(root->key < key){if(root->lchild == NULL){root->lchild = new tree(key);return;}else build(root->lchild, key);}else {if(root->rchild == NULL){root->rchild = new tree(key);return;}else build(root->rchild, key);}}void getPath(tree *root, int target){if(root->key == target){puts("");return;} else if(root->key > target){printf("E");getPath(root->rchild, target);}else{printf("W");getPath(root->lchild, target);} }int main(){int t, n, i, x, k;scanf("%d", &t);while(t--){scanf("%d", &n);scanf("%d", &x);tree *root = new tree(x);for(i = 1;i < n;i++){scanf("%d", &x);build(root, x);}scanf("%d", &k);for(i = 0;i < k;i++) scanf("%d", &q[i]);for(i = 0;i < k;i++) getPath(root, q[i]);}}

0 0