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push为前序遍历序列，pop为中序遍历序列。将题目转化为已知前序、中序，求后序。
前序GLR 中序LGR
前序第一个为G，在中序中找到G，左边为左子树L，右边为右子树R。
将左右子树看成新的树，同理。
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.
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.
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.
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
3 4 2 6 5 1
1 #include <iostream> 2 #include <cstdio> 3 #include <stack> 4 #include <string> 5 using namespace std; 6 7 #define MaxSize 30 8 9 #define OK 1 10 #define ERROR 0 11 12 int preOrder[MaxSize]; 13 int inOrder[MaxSize]; 14 int postOrder[MaxSize]; 15 16 void postorderTraversal(int preNo, int inNo, int postNo, int N); 17 18 int main() 19 { 20 stack<int> stack; 21 int N; //树的结点数 22 cin >> N; 23 string str; 24 int data; 25 int preNo = 0, inNo = 0, postNo = 0; 26 for(int i = 0; i < N * 2; i++) { //push + pop = N*2 27 cin >> str; 28 if(str == "Push") { //push为前序序列 29 cin >> data; 30 preOrder[preNo++] = data; 31 stack.push(data); 32 }else{ //pop出的是中序序列 33 inOrder[inNo++] = stack.top(); 34 stack.pop(); //pop() 移除栈顶元素（不会返回栈顶元素的值） 35 } 36 } 37 postorderTraversal(0, 0, 0, N); 38 for(int i = 0; i < N; i++) { //输出后序遍历序列 39 if(i == 0) //控制输出格式 40 printf("%d",postOrder[i]); 41 else 42 printf(" %d",postOrder[i]); 43 } 44 printf("\n"); 45 return 0; 46 } 47 48 void postorderTraversal(int preNo, int inNo, int postNo, int N) 49 { 50 if(N == 0) 51 return; 52 if(N == 1) { 53 postOrder[postNo] = preOrder[preNo]; 54 return; 55 } 56 int L, R; 57 int root = preOrder[preNo]; //先序遍历GLR第一个为根 58 postOrder[postNo + N -1] = root; //后序遍历LRG最后一个为根 59 for(int i = 0; i < N; i++) { 60 if(inOrder[inNo + i] == root) { //找到中序的根 左边为左子树 右边为右子树 61 L = i; //左子树的结点数 62 break; 63 } 64 } 65 R = N - L - 1; //右子树的结点数 66 postorderTraversal(preNo + 1, inNo, postNo, L); //同理，将左子树看成新的树 67 postorderTraversal(preNo + L + 1, inNo + L + 1, postNo + L, R);//同理，右子树 68 }
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原文地址：http://www.cnblogs.com/kuotian/p/5296968.html
