题目:
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
题目大意: 用栈的形式给出一棵二叉树的建立的顺序,求这棵二叉树的后序遍历
思路:根据给出的建立二叉树的栈操作,可以知道二叉树的前序遍历和中序遍历,从而就能够推导出二叉树的后序遍历。
如何得到前序遍历序列?每次Push操作的操作树按顺序插入数组,最终得到的数组就是前序遍历数组。
如何得到中序遍历序列?每次Pop操作从栈顶弹出的序号按顺序插入数组,最终得到的数组就是中序遍历数组。
由前序和中序遍历推到后序遍历的代码如下:(递归)
void postorder(int root, int start, int end) { // root是在前序遍历中根的下标 start 和 end是中序遍历的区间 if (start > end) return; int i = start; while (i < end && inOrder[i] != preOrder[root]) i++; postorder(root + 1, start, i - 1); // left postorder(root + 1 + i - start, i + 1, end); //right postOrder[post++] = preOrder[root]; }
代码:
#include<stdio.h> #include<iostream> #include<string.h> #include<stack> using namespace std; int preOrder[35], inOrder[35], postOrder[35]; int pre = 0, in = 0, post = 0; stack<int> s; int n; void postorder(int root, int start, int end) { if (start > end) return; int i = start; while (i < end && inOrder[i] != preOrder[root]) i++; postorder(root + 1, start, i - 1); postorder(root + 1 + i - start, i + 1, end); postOrder[post++] = preOrder[root]; } int main(){ scanf("%d", &n); for(int i = 0; i < 2 * n; i++){ string op; int x; cin>>op; if(op == "Push"){ cin>>x; s.push(x); preOrder[pre++] = x; }else if (op == "Pop"){ inOrder[in++] = s.top(); s.pop(); } } postorder(0, 0, n - 1); for(int i = 0; i < post; i++){ if(i != 0){ cout<<" "; } cout<<postOrder[i]; } return 0; }
标签:1086,遍历,end,int,Traversals,Tree,start,Push,root From: https://www.cnblogs.com/yccy/p/17416926.html