# 03-树3. Tree Traversals Again (25)

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题目实质是通过先序遍历和中序遍历建树，再后序遍历树。解题思路1. 通过输入建树    Push操作代表新建一个节点，将其与父节点连接并同时压栈    Pop操作，从栈顶弹出一个节点2. 后序遍历：递归实现代码如下：
#include <cstdio>
#include <cstring>
#include <cstdlib>

#define STR_LEN 5
#define MAX_SIZE 30

typedef struct Node
{
int data;
struct Node *left, *right;
}* treeNode;

treeNode Stack[MAX_SIZE];
int values[MAX_SIZE];

int num = 0;
int top = -1;

void Push(treeNode tn);
treeNode Pop();
treeNode Top();
bool isEmpty();

void PostOrderTraversal(treeNode root);

int main()
{
int n;
char operation[STR_LEN];
treeNode father, root;
bool findRoot = 0, Poped = 0;

scanf("%d", &n);
for (int i = 0; i < 2 * n; i++)
{
scanf("%s", operation);
if (strcmp(operation, "Push") == 0)
{
int value;
scanf("%d", &value);
treeNode newNode;
newNode = (treeNode)malloc(sizeof(struct Node));
newNode->data = value;
newNode->left = NULL;
newNode->right = NULL;
if (!findRoot)
{
root = newNode;     //根节点
Push(newNode);
findRoot = 1;
}
else
{
if (!Poped)     //如果前一个操作不是pop，则父节点为栈顶元素
father = Top();
if (father->left == NULL)
father->left = newNode;
else
father->right = newNode;
//printf("%d\n", newNode->data);
Push(newNode);
}
Poped = 0;
}
else
{
father = Pop();
Poped = 1;
}
}
PostOrderTraversal(root);

for (int i = 0; i < num-1; i++)
printf("%d ", values[i]);
printf("%d\n", values[num-1]);

return 0;
}

void PostOrderTraversal(treeNode root)
{
treeNode tn = root;
if(tn)
{
PostOrderTraversal(tn->left);
PostOrderTraversal(tn->right);
values[num++] = tn->data;       //将后序遍历出的节点值存入数组便于格式化打印
}
}

void Push(treeNode tn)
{
Stack[++top] = tn;
}

treeNode Pop()
{
return Stack[top--];
}

bool isEmpty()
{
}

treeNode Top()
{
return Stack[top];
}




posted @ 2015-08-21 19:05  llhthinker  阅读(4289)  评论(0编辑  收藏  举报
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