1086 Tree Traversals Again

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

题目大意：用栈来模拟以可二叉树的中序遍历，让你求出这棵二叉树的后序遍历。

大致思路：由题意可知二叉树的输入顺序为二叉树的前序遍历，然后我们借助一个栈来得出二叉树的后续遍历。进行一次Push操作我们就像栈中存一个元素，Pop操作我们就返回栈顶元素并将元素存入数组中。最后得到二叉树的中序遍历序列。

#include <bits/stdc++.h>

using namespace std;

const int N = 65;

struct TreeNode {

    int val;

    TreeNode* left;

    TreeNode* right;

    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

};

int n;

TreeNode root[N];

vector<int> preorder, inorder, postorder;  //先序序列，中序序列

void newNode(int x) {

    root[x].val = x;

    root[x].left = root[x].right = -1;

}

//给出了中序和先序构建二叉树

TreeNode* create(int preL, int preR, int inL, int inR) {

    if (preL > preR) return -1;

    TreeNode* root = new TreeNode(preorder[preL]);

    int k;

    for (k = inL; k <= inR; k++) {

        if (inorder[k] == preorder[preL]) break;

    }

    int numL = k - inL;

    root->left = create(preL + 1, preL + numL, inL, k - 1);

    root->right = create(preL + numL + 1, preR, k + 1, inR);

    return root;

}

void postvisit(TreeNode* root) {

    if (node == -1) return;

    postvisit(root->left);

    postvisit(root->right);

    postorder.push_back(root->val);

}

int main() {

    scanf("%d", &n);

    //先将每个节点初始化

    stack<int> tmp;

    string op;

    int id;

    while(n--) {

        cin >> op;

        if (op == "Push") {

            cin >> id;

            tmp.push(id);

            preorder.push_back(id);

        }

        else {

            inorder.push_back(tmp.top());

            tmp.pop();

        }

    }

    create(0, n - 1, 0, n - 1);

    TreeNode* root = create(0, n - 1, 0 , n - 1);

    postvisit(root);

    for (int i = 0; i < postorder.size(); i++) {

        cout << postorder[i];

        if (i != postorder.size() - 1) cout << " ";

        else cout << endl;

    }

    return 0;

}

巴特西

1086 Tree Traversals Again——PAT甲级真题

1086 Tree Traversals Again

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