03-树3. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1

#include <stdio.h>struct Node{int tag;//标记节点是第几次进栈int num;};//先序遍历对应进栈顺序，中序遍历对应出栈顺序；//后序遍历与中序遍历不同的是节点出栈后要马上再入栈（tag做第二次入栈标记），等右儿子遍历完后再出栈;//具体实现上，每次中序遍历的pop时，如果栈顶是标记过的节点（tag=2），循环弹出；如果没有标记过（tag=1），做标记，即弹出再压栈）//栈顶tag=2的节点对应中序遍历中已弹出的节点；循环弹出后碰到的第一个tag=1的节点才对应中序遍历当前pop的节点int main() {freopen("test.txt", "r", stdin);int n;scanf("%d", &n);int flag = 0;struct Node stack[30];int size = 0;//栈元素大小，指向栈顶的下一个位置for (int i = 0; i < 2 * n; ++i) {char s[10];scanf("%s", s);if (s[1] == 'u') {//pushscanf("%d", &stack[size].num);//入栈stack[size].tag = 1;//标记第一次入栈++size;}else {//popwhile (size > 0 && stack[size - 1].tag == 2) {//循环弹出栈顶tag=2的节点if (flag)printf(" ");flag = 1;printf("%d", stack[--size].num);}if (size > 0)//将中序遍历中应该要弹出的节点弹出再压栈，做标记即可stack[size - 1].tag = 2;}}while (size) {//将栈中剩余节点依次弹出if (flag) {printf(" ");}flag = 1;printf("%d", stack[--size].num);}return 0;}

题目链接：http://www.patest.cn/contests/mooc-ds/03-%E6%A0%913