PAT(Advanced Level)A1138. Postorder Traversal

题意

假设二叉树里面的数字都是正数且各不相同，给先序和中序遍历， 要输出后序遍历的第一个元素

思路

• 先序和中序遍历序列可以唯一确定一颗二叉树，我们可以建树之后后序遍历就好

代码

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;
struct node {
    int val;
    struct node* left;
    struct node* right;
    node(int x): val(x), left(NULL), right(NULL){}
};
unordered_map<int, int> lookup;
node* build_tree(vector<int>& in_order, int inL, int inR, vector<int>& pre_order, int preL, int preR) {
    if(inL > inR) {
        return NULL;
    }
    node *root = new node(pre_order[preL]);
    int pos = lookup[root->val];
    int num_left = pos - inL;
    root->left = build_tree(in_order, inL, pos - 1, pre_order, preL + 1, preL + num_left);
    root->right = build_tree(in_order, pos + 1, inR, pre_order, preL + num_left + 1, preR);
    return root;
}
void post_travel(node* cur, vector<int>& post_order) {
    if(cur) {
        post_travel(cur->left, post_order);
        post_travel(cur->right, post_order);
        post_order.emplace_back(cur->val);
    }
}
int main() {
    vector<int> pre_order, in_order;
    int N;
    scanf("%d", &N);
    pre_order.resize(N);
    in_order.resize(N);
    for(int i = 0; i < N; i++)  scanf("%d", &pre_order[i]);
    for(int i = 0; i < N; i++) {
        scanf("%d", &in_order[i]);
        lookup[in_order[i]] = i;
    }
    
    node* root = build_tree(in_order, 0, N - 1, pre_order, 0, N - 1);
    
    vector<int> post_order;
    post_travel(root, post_order);
    cout << post_order.front();
    return 0;
}