An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
Figure 1
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: “Push X” where X is the index of the node being pushed onto the stack; or “Pop” meaning to pop one node from the stack.
Output Specification:
For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1
这道题就是一道已知先序和中序然后让你用后序遍历树并输出,push的顺序即为先序,pop的顺序即为中序。那么就需要递归的寻找根节点,先序的第一个元素为根节点,那么中序中找到根节点,根节点的前半部分为左子树,后半部分为右子树,就这样不断递归,然后存入后序中即可,详见代码及注释
比如题目中的例子
先序:1 2 3 4 5 6
中序:3 2 4 1 6 5
后序:3 4 2 6 5 1
#include <iostream>
#include <string>
#include <stack>
#include <vector>
using namespace std;
void GetPostOrder(vector<int> &PreOrder/*先序数组*/,int PreL/*先序指针*/,vector<int> &InOrder/*中序数组*/,int InL/*中序指针*/,
vector<int> &PostOrder/*后序数组*/,int PostL/*后序指针*/,int n)
{
if(n == 0)
{
return;
}
else if( n == 1)
{
PostOrder[PostL] = PreOrder[PreL];
return;
}
int root = PreOrder[PreL];/*先序序列的第一个为根节点*/
PostOrder[PostL + n - 1] = root;/*后序序列中最后访问根节点*/
/*在中序序列中找到根节点*/
int i = 0;
while(i < n)
{
if(InOrder[InL + i] == root)
{
break;/*找到对应根节点*/
}
i++;
}
int L = i;/*左子树*/
int R = n - i - 1;/*右子树*/
GetPostOrder(PreOrder,PreL + 1,InOrder,InL,PostOrder,PostL,L);/*遍历左子树*/
GetPostOrder(PreOrder,PreL + L + 1,InOrder,InL + L + 1,PostOrder,PostL + L,R);/*遍历右子树*/
}
int main()
{
int n;
cin >> n;
vector<int> PreOrder(n,0);
vector<int> InOrder(n,0);
stack<int> st;
int PreL = 0,InL = 0;
for(int i = 0 ; i < 2 * n; i++)
{
string str;
int temp;
cin >> str;
if(str == "Push")
{
cin >> temp;
PreOrder[PreL++] = temp;/*先序序列*/
st.push(temp);
}
else
{
InOrder[InL++] = st.top();/*中序序列*/
st.pop();
}
}
vector<int> PostOrder(n,0);
GetPostOrder(PreOrder,0,InOrder,0,PostOrder,0,n);
for(int i = 0 ; i < n ; i++)
{
if(i < n - 1)
{
cout << PostOrder[i] << " ";
}
else
{
cout << PostOrder[i] << endl;
}
}
return 0;
}