PAT 1086 Tree Traversals Again

室友在做于是也做一发,跟已知两种遍历序列还原二叉树的思路类似,感觉PAT上的题目跟书本知识靠的近一些

#include <iostream>
#include <cstdio>
#include <vector>using namespace std;void print_data(const vector<char> &as, const vector<int> &ns) {int len = ns.size();for (int i=0; i<len; i++) {if (as[i] == 'i') {cout<<"push "<<ns[i]<<endl;} else {cout<<"pop"<<endl;}}
}void build_postorder(vector<char> &actions, vector<int> &nums, int start, int end, vector<int> &postorder) {if (start >= end) return;if (actions[start] != 'i') return;// root node
    postorder.push_back(nums[start]);// now try to find the seperator idx of // action(must be a pop pair wise with root node push) between two sub treeint pushs = 1;int idx = start + 1;while (pushs != 0 && idx < end) {if (actions[idx] == 'i') {pushs++;} else {pushs--;}idx++;}// right sub tree
    build_postorder(actions, nums, idx, end, postorder);// left sub treebuild_postorder(actions, nums, start + 1, idx, postorder);
}
int main() {int N, len;scanf("%d", &N);len = N * 2;char buf[10];int num;vector<char> actions(2 * N, '\0');vector<int> nums(2 * N, 0);// read in datafor (int i=0; i<len; i++) {scanf("%s", buf);if (buf[3] == 'h') {scanf("%d", &num);nums[i] = num;actions[i] = 'i';    // push, input} else {actions[i] = 'o';    // pop, output
        }}vector<int> postorder;build_postorder(actions, nums, 0, len, postorder);if (postorder.size() > 0){// print_data(actions, nums);for (int i=N - 1; i>=1; i--) {printf("%d ", postorder[i]);}printf("%d", postorder[0]);}return 0;
}

转载于:https://www.cnblogs.com/lailailai/p/3967525.html

PAT 1086 Tree Traversals Again相关推荐

PAT甲级1086 Tree Traversals Again：[C++题解]二叉树中序序列、栈、求后序遍历
文章目录题目分析题目链接题目分析分析: 给定栈模拟的二叉树的中序序列. 我们可以发现一些性质: 1 第一个值是根结点. 2 对于所有的push操作,如果上一个是push,该结点就是上一个结点的 ...
1086 Tree Traversals Again (25分)
1 题目 1086 Tree Traversals Again (25分) An inorder binary tree traversal can be implemented in a non-r ...
1086 Tree Traversals Again
1. 这题的核心部分是,根据二叉树的先序序列和中序序列求后序序列.等于是在1020 Tree Traversals这一题的基础上,把怎么得到先序序列和中序序列的难度加大了,不是直接给出,而是要曲折一点 ...
【PAT (Advanced Level) Practice】1086 Tree Traversals Again (25 分)
众所周知只有前序遍历和后序遍历是不可能推出中序遍历的,所以我就不废话了由树的前序遍历和中序遍历可以推后序遍历由树的中序遍历和后序遍历可以推前序遍历 #include <cstdio> ...
PAT 1020. Tree Traversals
题目:http://pat.zju.edu.cn/contests/pat-a-practise/1020 题解: 题意就是给一个二叉树的后序和中序,然后确定这棵二叉树,然后层序输出这棵树. 主要思路 ...
1086 Tree Traversals Again (25 分)【一般 / 建树树的遍历】
https://pintia.cn/problem-sets/994805342720868352/problems/994805380754817024 首先要知道的是: 入栈顺序就是树的前序遍历的 ...
PAT甲级1020 Tree Traversals：[C++题解]树的遍历、由中序序列和后序序列递归建树
文章目录题目分析题目链接题目分析题意重述:给定一棵二叉树的后序遍历序列和中序遍历序列,让求层次遍历的序列. 分析: 后序遍历:先左子树.右子树 ,最后再遍历根结点. 中序遍历:先左子树,再根 ...
1020. Tree Traversals (25)
题目链接:http://www.patest.cn/contests/pat-a-practise/1020 题目: 1020. Tree Traversals (25) 时间限制 400 ms 内存 ...
03-树3 Tree Traversals Again
题目 An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For exam ...
hdu 1710 Binary Tree Traversals (二叉树）
1 /********************************************************** 2 题目: Binary Tree Traversals(hdu 1710) ...

PAT 1086 Tree Traversals Again

PAT 1086 Tree Traversals Again相关推荐

最新文章

热门文章