1086 Tree Traversals Again✍个人博客:https://blog.csdn.net/Newin2020?spm=1011.2415.3001.5343
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📝原题地址:题目详情 - 1086 Tree Traversals Again (pintia.cn)
🔑中文翻译:再次树遍历
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题意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:
Figure 1Each 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:Sample Output:6 Push 1 Push 2 Push 3 Pop Pop Push 4 Pop Pop Push 5 Push 6 Pop Pop
3 4 2 6 5 1
通过使用栈可以以非递归方式实现二叉树的中序遍历。
例如,假设遍历一个如下图所示的 6 6 6 节点的二叉树(节点编号从 1 1 1 到 6 6 6)。
则堆栈 *** 作为:push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop()
。
我们可以从此 *** 作序列中生成唯一的二叉树。
你的任务是给出这棵树的后序遍历。
思路可以从用栈实现二叉树的中序遍历的思路中找到规律:
- 如果上一次 *** 作是
Push
,则当前Push
的结点是其父结点的左孩子,这里的父结点是上次Push
的结点。例如,push(1); push(2);
,则结点2
是上次Push
的结点1
的左孩子。 - 如果上一次 *** 作是
Pop
,则当前Push
的结点是其父结点的右孩子,这里的父节点是上次Pop
的结点。例如,push(1); push(2); push(3); pop(); pop(); push(4);
,结点4
是上次Pop
的结点2
的右孩子。 - 如果上一次没有 *** 作,则当前
Push
的结点就是根结点。
#include
using namespace std;
const int N = 40;
int n;
int l[N], r[N]; //用哈希表来存储每个结点左孩子的下标
void dfs(int u, int root)
{
if (!u) return;
dfs(l[u], root);
dfs(r[u], root);
cout << u;
if (u != root) cout << " "; //最后一个结点输出末尾不能加空格
}
int main()
{
int last = 0, type, root;
stack<int> stk;
cin >> n;
for (int i = 0; i < n * 2; i++)
{
string op;
cin >> op;
if (op == "Push")
{
int x;
cin >> x;
if (!last) root = x; //第一个结点为根结点
else //设置父结点的左右孩子
{
if (type == 0) l[last] = x; //上个 *** 作是Push
else r[last] = x; //上个 *** 作是Pop
}
last = x;
type = 0; //表示Push
stk.push(x);
}
else
{
last = stk.top();
stk.pop();
type = 1; //表示Pop
}
}
dfs(root, root); //后序遍历
return 0;
}
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