【PAT】1086. 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

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.youkuaiyun.com/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;
}