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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
题意解析：入栈顺序即为先序遍历的顺序，出栈顺序即为中序遍历的顺序
解题思路：
1.根据入栈出栈顺序，建立先序遍历数组与中序遍历数组。
2.取先序序列中的第一个元素，该元素为根结点
3.根据根结点在中序序列中查找根结点的位置，从而得到该树左子树结点个数（L）与右子树的结点个数（R）
4.在后序序列数组中，第0到第L个元素为左子树，第L+1到第L+R个元素为右子树，最后一个元素为根结点
#include <iostream> #include <string> #include <stack> using namespace std; #define MAXSIZE 30 int preOrder[MAXSIZE]; int inOrder[MAXSIZE]; int postOrder[MAXSIZE]; void Solve(int preL, int inL, int postL, int n); int main() { for (int i = 0; i < MAXSIZE; i++) //初始化数组 { preOrder[i] = 0; inOrder[i] = 0; postOrder[i] = 0; } stack<int> inputStack; int nodeNum; cin >> nodeNum; int i, data; int preIndex = 0, inIndex = 0, postIndex = 0; string str; for (i = 0; i < 2 * nodeNum; i++) { cin >> str; if (str == "Push") { cin >> data; preOrder[preIndex++] = data; inputStack.push(data); } else if (str == "Pop") { inOrder[inIndex++] = inputStack.top(); inputStack.pop(); } } Solve(0, 0, 0, nodeNum); for (int i = 0; i < nodeNum; i++) { if ( i == nodeNum - 1 ) { cout << postOrder[i] << endl; } else { cout << postOrder[i] << ‘ ‘; } } } void Solve(int preL, int inL, int postL, int n) { if ( n == 0 ) { return; } if ( n == 1 ) { postOrder[postL] = preOrder[preL]; return; } int root = preOrder[preL]; postOrder[postL + n - 1] = root; int L, R; int i; for (i = 0; i < n; i++) { if ( inOrder[inL + i] == root ) { break; } } L = i; //左子树结点个数 R = n - L - 1; //右子树结点个数 Solve(preL + 1, inL, postL, L); Solve(preL + L + 1, inL + L + 1, postL + L, R); }
03-3. Tree Traversals Again (PAT)
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原文地址：http://www.cnblogs.com/liangchao/p/4281458.html
