03-树3. Tree Traversals Again

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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.

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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.

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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

代码如下：

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#include<iostream>
#include<stack>
#include<vector>
using namespace std;

vector<int> Tree[30];
vector<int> ans;
void postOrderTraversal(int fNode)  //fNode : father node
{
    if(Tree[fNode].size() == 1)
    {
        postOrderTraversal(Tree[fNode][0]);
        ans.push_back(fNode);
    }
    else if(Tree[fNode].size() == 2)
    {
        postOrderTraversal(Tree[fNode][0]);
        postOrderTraversal(Tree[fNode][1]);
        ans.push_back(fNode);
    }
    else
    {
        ans.push_back(fNode);
    }
}
int main()
{
    int N = 0;
    cin >> N;

    string str;
    int val;
    stack<int> stk;
    int popVal = -1;
    int root = 0;
    for(int i = 0; i < N * 2; ++i)
    {
        cin >> str;
        if(str == "Push")
        {
            cin >> val;
            if(i != 0 && popVal == -1)
            {
                Tree[stk.top()].push_back(val);
            }
            else if(i == 0)
            {
                root = val;
            }
            if(popVal != -1)
            {
                Tree[popVal].push_back(val);
            }
            stk.push(val);
            popVal = -1;
        }
        else if(str == "Pop")
        {
            popVal = stk.top();
            stk.pop();
        }
    }
    postOrderTraversal(root);
    for(int i = 0; i < ans.size(); ++i)
    {
        cout << ans[i] << ((i == ans.size() - 1)?"\n":" ");
    }
    return 0;
}

这是我第二次一遍就能AC的代码！
思路：
push即为先序遍历序列
pop即为中序遍历序列
知道这两个序列，就可以唯一确定一棵树，然后在后序遍历即可。
如何分辨push进来的元素是哪个结点（父结点）的左子结点还是右子结点？
每次push之前，如果之前那次是pop操作，则父结点为pop的那个元素，如果之前依然是push操作，那么父结点就是栈顶元素。代码中设置一个popVal来判断上一次操作是pop还是push，是pop，记录pop元素的值，以便接下来push操作找父结点。