PAT甲級1086 Tree Traversals Again//中序+前序組建二叉樹
技術標籤:# PAT甲級
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.
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
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1
思路
可以先看這題PAT甲級1020 Tree Traversals//由後序和中序序列確定一棵二叉樹
一個棧可以模擬一棵二叉樹的先序和中序遍歷過程,比如題目中樣例的操作:
Push的次序是:1、2、3、4、5、6,所以先序序列就是123456
Pop的次序是:3、2、4、1、6、5,所以中序序列就是324165
可以根據先序+中序得出一棵樹,再得出後序序列。
思路就是分別提取先序和中序,然後建造一棵樹,最後得出後序序列。
- non-recursive 非遞迴的
#include <iostream>
#include <cstdlib>
#include <vector>
#include <string>
using namespace std;
struct node
{
int val;
node* left;
node* right;
};
vector<int> pre;
vector<int> in;
vector<int> ans;
int stack[61];
node* creat(int preL, int preR, int inL, int inR);
void post(node*);
int main()
{
int n, index = 0;
cin >> n;
getchar();
for (int i = 0; i < 2 * n; i++)
{
string op;
getline(cin, op);
if (op[1] == 'u')
{
int num = stoi(op.substr(5));
stack[index++] = num;
pre.push_back(num);
}
else
{
int num = stack[--index];
stack[index] = 0;
in.push_back(num);
}
}
node* root = creat(0, pre.size() -1, 0, in.size() - 1);
post(root);
for (int i = 0; i < ans.size(); i++)
{
if (i > 0)
cout << " ";
cout << ans[i];
}
system("pause");
return 0;
}
node* creat(int preL, int preR, int inL, int inR)
{
if (preL > preR)
return NULL;
node* root = new node;
root->val = pre[preL];
int k = 0;
for (int i = 0; i < in.size(); i++)
{
if (in[i] == pre[preL])
{
k = i;
break;
}
}
int num_left = k - inL;
root->left = creat(preL + 1, preL + num_left, inL, k - 1);
root->right = creat(preL + num_left + 1, preR, k + 1, inR);
return root;
}
void post(node* root)
{
if (!root)
return;
post(root->left);
post(root->right);
ans.push_back(root->val);
}