1138 Postorder Traversal

题目：

Suppose that all the keys in a binary tree are distinct positive integers. Given the preorder and inorder traversal sequences, you are supposed to output the first number of the postorder traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤50,000), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the first number of the postorder traversal sequence of the corresponding binary tree.

Sample Input:

7
1 2 3 4 5 6 7
2 3 1 5 4 7 6

Sample Output:

3

 

优化：先根据中序和前序遍历构建树，之后再进行后序遍历，优化后可以将上面两部操作合并，详见如下代码

 

代码：

#include <iostream>
#include <vector>
using namespace std;
vector<int> pre, in;
bool flag = false;
void postOrder(int prel, int inl, int inr) {
    if (inl > inr || flag == true) return;
    int i = inl;
    while (in[i] != pre[prel]) i++;
    postOrder(prel+1, inl, i-1);
    postOrder(prel+i-inl+1, i+1, inr);
    if (flag == false) {
        printf("%d", in[i]);
        flag = true;
    }
}
int main() {
    int n;
    scanf("%d", &n);
    pre.resize(n);
    in.resize(n);
    for (int i = 0; i < n; i++) scanf("%d", &pre[i]);
    for (int i = 0; i < n; i++) scanf("%d", &in[i]);
    postOrder(0, 0, n-1);
    return 0;
}