POJ 2828 Buy Tickets
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树的中序遍历,具体思路见代码。。。
#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>using namespace std;struct N{ int ans,data;//ans存贮的为该节点的左子树有 ans-1 个节点 N *l,*r; //即该节点为 以该节点为根节点的树中的位置};N *creat(){ N *root = (N *)malloc(sizeof(N)); root->l = root->r = NULL; return root;}void insert(int site,int data,N *&root){ if(root == NULL) { root = creat(); root->ans = 1; root->data = data; return ; } if(site < root->ans) { root->ans++; insert(site,data,root->l);//若site < ans则说明要寻找的点在左子树中 } else { insert(site-root->ans,data,root->r); //因为此节点的右子树中有 ans-1 个节点 //所以只需在右子树中寻找 第 site-ans 个节点 }}bool MarkBlance;void output(N *root)//树的中序遍历{ if(root == NULL) { return ; } output(root->l); if(MarkBlance == true) { printf("%d",root->data); MarkBlance = false; } else { printf(" %d",root->data); } output(root->r);}int main(){ int n; int site,data; N *root ; while(scanf("%d",&n) != EOF) { root = NULL; while(n--) { scanf("%d %d",&site,&data); insert(site,data,root); } MarkBlance = true;//控制空格 output(root); cout<<endl; } return 0;}
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