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.
  

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

分析思路

• Push操作刚好是树的前序遍历过程；Pop操作刚好是树的中序遍历操作

AC代码

/*!* \file 03-树3 Tree Traversals Again.cpp** \author ranjiewen* \date 2017/04/25 23:51** */#include <iostream>
#include <cstdio>
#include <stack>
#include <string>using namespace std;#define MaxSive 30
#define OK 0
#define ERROR -1int preOrder[MaxSive];
int inOrder[MaxSive];
int postOrder[MaxSive];void postOrderTraversal(int preNum,int inNum,int postNum,int Num);int main()
{stack<int> s;int N; //树的结点数cin >> N;string str;int data;int preNum = 0, inNum = 0, postNum = 0;for (int i = 0; i < N * 2;i++)  //push+pop=N*2 {cin >> str;if (str=="Push") //Push为前序序列{cin >> data;preOrder[preNum++] = data;s.push(data);}else  //Pop为中序序列{inOrder[inNum++] = s.top();s.pop();  //移除栈顶元素（不会返回栈顶元素的值）}}postOrderTraversal(0,0,0,N); //每棵子树前，中，后序数组的开始下标，N为子树的结点个数for (int i = 0; i < N;i++)  //输出后序遍历序列{if (i==0)   //输出格式控制{printf("%d", postOrder[i]);}else{printf(" %d", postOrder[i]);}}printf("\n");return 0;
}void postOrderTraversal(int preNum, int inNum, int postNum, int Num)
{if (Num==0){return;}if (Num==1){postOrder[postNum] = preOrder[preNum];return;}int L=0, R=0; //递归左右子树的结点个数int root = preOrder[preNum]; //先序遍历的第一个节点为根节点postOrder[postNum + Num - 1] = root; //填后序遍历的坑for (int i = 0; i < Num;i++){if (inOrder[i+inNum] == root)//不要掉了preNum //在中序遍历中找到根节点，分为左右子树递归{L = i;  //左子树结点个数break;}}R = Num- L - 1; //右子树结点个数postOrderTraversal(preNum+1,inNum,postNum,L);postOrderTraversal(preNum + L + 1, inNum + L + 1, postNum + L, R); //右子树递归调用，注意开始下标}

Reference

题目来源