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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1

入栈顺序为前序遍历，出栈顺序为中序遍历，根据前序遍历序列和中序遍历序列确定二叉树.也可以根据输入递归地建树

#include <iostream>#include <string>#include <vector>using namespace std;int n;struct node{int left,right;node(int l = -1,int r = -1):left(l),right(r){}};vector<node> tree;int inorder(){string s;int t;cin>>s;if(s[1] == 'u'){cin>>t;tree[t].left = inorder();tree[t].right = inorder();return t;}else return -1;}int kase = 0;void posterorder(int r){if(r != -1){posterorder(tree[r].left);posterorder(tree[r].right);if(kase++)cout<<" ";cout<<r;}}int main(){cin>>n;tree.resize(n+1);posterorder(inorder());return 0;}