03-树3 Tree Traversals Again（25 point(s)）
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.
Figure 1 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 Pop Push 4 Pop Pop Push 5 Push 6 Pop Pop
Sample Output:
3 4 2 6 5 1
思路
题目给出 入栈的过程 要求求出 该树 的后序遍历结果
入栈过程 其实就是 前序遍历的结果
然后出栈过程 是中序遍历结果
所以实际上 题目的意思 就是 给出 前序遍历 和 中序遍历 求出 后序遍历
根据题给的样例
前序遍历 1 2 3 4 5 6 中序遍历 3 2 4 1 6 5
前序遍历 是 根 左 右 中序遍历 是 左 根 右 后序遍历 是 左 右 根
所以 我们可以认为 对于一棵树 来说 前序遍历的 第一个节点 就是它的根节点
然后 从 中序遍历 去找 直到 找到根节点，那么 根节点 前面的元素 就是左子树 右边的 元素 就是 右子树 然后 再去 前序遍历 找相应的元素区间 就是 对应的 左子树 区间 和 右子树 区间
其实下面的过程 就是 递归 重叠子问题的过程
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; typedef pair<string, int> psi; typedef pair<string, string> pss; const double PI = 3.14159265358979323846264338327; const double E = exp(1); const double eps = 1e-30; const int INF = 0x3f3f3f3f; const int maxn = 1e3 + 5; const int MOD = 1e9 + 7; vector <int> pre, in, ans; void post(int root, int start, int end) { if (start >= end) return; int i = start; while (i < end && in[i] != pre[root]) i++; int len = i - start; post(root + 1, start, i); post(root + len + 1, i + 1, end); ans.pb(pre[root]); } void print(vector <int> v) { vector <int>::iterator it; for (it = v.begin(); it != v.end(); it++) { if (it != v.begin()) printf(" "); printf("%d", (*it)); } printf("\n"); } int main() { stack <int> st; int n; scanf("%d", &n); int m = n << 1; string s; int num; for (int i = 0; i < m; i++) { cin >> s; if (s == "Push") { scanf("%d", &num); st.push(num); pre.pb(num); } else { in.pb(st.top()); st.pop(); } } post(0, 0, n); print(ans); }转载于:https://www.cnblogs.com/Dup4/p/9433165.html
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