03-树3 Tree Traversals Again (25 分)

03-树3 Tree Traversals Again (25 分)

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

思路

1. 树的遍历中，除根节点以外其他节点每个结点被访问三次，push一次，pop一次，pop一次因而题目转换成了已知先序和中序求后序
2. 具体思路为每次递归求出子树的根节点
3. vector 类型的数组 参数从 int a[] 变成 vector < int > a

例子

pre : 1 2 3 4 5 6
mid : 3 2 4 1 6 5
post: 3 4 2 6 5 1

Codes

#include
#include
#include
#include
using namespace std;
void getPostOrder(vector pre, int preL, vector mid, int midL,vector &post, int postL, int n) //preL midL postl is pointer
{
    if (n == 0) return ;
    if (n == 1) {
        post[postL] = pre[preL];
        return ;
    }
    int root = pre[preL];
    post[postL + n - 1] = root;
    int i = 0;
    while (i < n) {
        if (mid[midL + i] == root)  break;
        i++;
    }
    int L = i, nodesnum = n - i - 1;
	getPostOrder(pre, preL + 1, mid, midL, post, postL, L);
    getPostOrder(pre, preL + L + 1, mid, midL + L + 1, post, postL + L, nodesnum);
}

int main() {
    int num;
    cin >> num;
    vector  pre(num, 0) , mid(num, 0), post(num, 0);
    stack s;
    int i = 0;
    for(int cnt = 0; i < num; ){
        string temp ;
        int tempdata;
        cin >> temp;
        if(temp == "Push"){
			cin >> tempdata;
            pre[cnt] = tempdata;
            s.push(tempdata);
            cnt++;
        }
        else{
            mid[i++] = s.top();
            s.pop();
        }
    }
    getPostOrder(pre, 0,   mid, 0 , post, 0 , num);
    for(int i = 0; i < num; i++) {
        if(i < num -1) cout << post[i] << " ";
        else cout << post[i];
    }
    return 0;
}
 

Code2 dfs

#include
using namespace std;
vector res;
int num;
int a[10010] , intern[10010];
int cnt = 0;
int ready = 0;
void Dfs(int n){
	if(ready == 2) return;
	int tempdata;
	string temp;	
	cin >> temp;
	if(temp == "Push"){
		cin >> tempdata;
		a[cnt++] = tempdata;

	}else {
	    if(cnt == num){
			ready ++;
		}
        return;
	}
	Dfs(n * 2 + 1);
	Dfs(n * 2 + 2);
	res.push_back(tempdata);
}
int main(){
	cin >> num;
	Dfs(0);
	for(int i = 0; i < num; i++){
		if( i== 0) cout << res[i];
		else cout << " " << res[i];
	}
	return 0;
}