1138 Postorder Traversal（25 分）（C++）

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本文介绍了一种算法，该算法通过给定的先序和中序遍历序列，求解二叉树的后序遍历的第一个节点。具体实现包括输入二叉树节点数、先序和中序遍历序列，然后通过递归方式构建二叉树并输出后序遍历的第一个节点。

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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:

已知一个二叉树的先序和中序遍历，求其后序遍历的第一个节点。

#include <iostream>
#include <vector>
using namespace std;
int n, ans = -1;
vector<int>pre, in;
void posttravel(int inleft, int inright, int preleft){
	if(ans != -1 || inleft > inright)
		return;
	int i = inleft;
	while(in[i] != pre[preleft])
		++ i;
	posttravel(inleft, i - 1, preleft + 1);
	posttravel(i + 1, inright, preleft + i - inleft + 1);
	if(ans == -1)
		ans = pre[preleft];
}
int main(){
	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]);
	posttravel(0, n - 1, 0);
	printf("%d", ans);
}