# 1086. Tree Traversals Again (25)-PAT甲级真题（树的遍历，前序中序转后序）

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

root为当前子树的根结点在前序pre中的下标，start和end为当前子树的最左边和最右边的结点在中序in中的下标。用i找到当前子树的根结点root在中序中的下标，然后左边和右边就分别为当前根结点root的左子树和右子树。递归实现～
Update：Github用户littlesevenmo给我发issue提出题目并没有说所有节点的值互不相同。因此，在有多个节点的值相同的情况下，之前的代码会输出错误的结果，所以修改后的代码中添加了key作为索引，前中后序中均保存索引值，然后用value存储具体的值，修改后的代码如下：

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