Binary Search Tree（二叉搜索树）

Problem description

A binary search tree is a binary tree that satisfies the following properties:

• The left subtree of a node contains only nodes with keys less than the node’s key.

• The right subtree of a node contains only nodes with keys greater than the node’s key.

• Both the left and right subtrees must also be binary search trees.



Pre-order traversal (Root-Left-Right) prints out the nodes key by visiting the root node then traversing the left subtree and then traversing the right subtree. Post-order traversal (Left Right-Root) prints out the left subtree first and then right subtree and finally the root node. For example, the results of pre-order traversal and post-order traversal of the binary tree shown in Figure 1 are as follows: Pre-order: 50 30 24 5 28 45 98 52 60 Post-order: 5 28 24 45 30 60 52 98 50 Given the pre-order traversal of a binary search tree, you task is to find the post-order traversal of this tree.

Input

The keys of all nodes of the input binary search tree are given according to pre-order traversal. Each node has a key value which is a positive integer less than 106. All values are given in separate lines (one integer per line). You can assume that a binary search tree does not contain more than 10,000 nodes and there are no duplicate nodes.

Output

The output contains the result of post-order traversal of the input binary tree. Print out one key per line.

Sample Input

50

30

24

5

28

45

98

52

60

Sample Output

5

28

24

45

30

60

52

98

50

题意

给一个序列，让它以二叉树排序树的形式插入，即:遇到值比他小的往左子树插入，否则往右子树插入，输出他的后续遍历序列。

思路

建议大家能用数组的时候尽量开固定大小的数组，因为用指针去写的话申请内存耗时间，而且有的题目还可能内存超限，情况比较少就是了。这里先给出指针形式再给出数组形式代码。





图一是数组方法，图二是指针方法，比较两者时间，可想而知。

方法一

#include<iostream>
#include<stdio.h>
using namespace std;
struct tree
{
int data;
tree *l=NULL,*r=NULL;
};
void _insert(tree *&T,int e)
{
if(!T)
{
tree *S=new tree;
S->data=e;
T=S;
}else if(e<T->data){
_insert(T->l,e);
}else{
_insert(T->r,e);
}
}
void _print(tree *&T)
{
if(T){
_print(T->l);
_print(T->r);
printf("%d\n",T->data);
}
}
int main()
{
int e;
tree *T=NULL;
while(~scanf("%d",&e)){
_insert(T,e);
}
_print(T);
return 0;
}

方法二

#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn = 1e6 + 10;
using namespace std;

int son[maxn][2];
int n, root = 0;

void __insert(int val, int &x) {
if(x==0){
x=val;
return ;
}
if(val < x)
__insert(val, son[x][0]);
else
__insert(val, son[x][1]);
}

void dfs(int o) {
if(o==0)
return ;
dfs(son[o][0]);
dfs(son[o][1]);
printf("%d\n", o);
}

int main() {
while(scanf("%d", &n) != EOF)
__insert(n, root);
dfs(root);
return 0;
}

练习网址：https://cn.vjudge.net/problem/UVA-12347