【PAT】1086. Tree Traversals Again (25)

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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:

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

Sample Output:

3 4 2 6 5 1

分析：根据题目给出的push，pop操作可以得出前序数组和中序数组，从而可以构建树，然后再进行后序遍历。

(1) 题目中的push操作对应的数组就是前序数组， (1,2,3,4,5,6)

(2) 题目中的pop操作对应的数组就是中序数组，（3,2,4,1,6,5）

(3) 前序数组的第一个元素就是树的根1。而在中序数组中，数字1的左边部分就是以1为根的左子树中的节点，数字1的右边部分就是以1为根的右子树中的节点。

建议参考： http://blog.csdn.net/solin205/article/details/39157781

正确代码如下：

#include <iostream>#include <vector>#include <string>#include <stack>using namespace std;struct Node{Node *left, *right;int val;};Node * buildTree(int *preorder, int *inorder, int length){if(length == 0)   return NULL;int val = preorder[0];int i, index;for(i=0; i<length; i++){if(inorder[i] == val){index = i;break;}}Node *N = new Node();N->val = val;N->left = buildTree(preorder+1, inorder, index);N->right = buildTree(preorder+index+1, inorder+index+1, length-index-1);return N;}void buildPostorder(Node *root, vector<int> &postorder){if(root == NULL)return;buildPostorder(root->left, postorder);buildPostorder(root->right, postorder);postorder.push_back(root->val);}int main(){int n, i, val;scanf("%d",&n);string op;int preorder[50], inorder[50];int  preIndex=0, inIndex=0;stack<int> sta;for(i=0; i<2*n; i++){cin>>op;if(op == "Push"){scanf("%d", &val);sta.push(val);preorder[preIndex++] = val;}else if(op == "Pop"){inorder[inIndex++] = sta.top();sta.pop();}}Node *root;vector<int> postorder;root = buildTree(preorder, inorder, n);buildPostorder(root, postorder);for(i=0; i<postorder.size(); i++){if(i==0)  printf("%d",postorder[i]);else  printf(" %d",postorder[i]);}printf("\n");return 0;}

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