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
思路分析：
在示例中
Push入栈的顺序是 1，2，3，4，5，6，是访问二叉树的先序遍历
Pop出栈顺序是 3，2，4，1，6，5，是中序遍历二叉树
于是问题变成已知先序遍历 中序遍历，求二叉树的后序遍历
AC代码如下
#include<cstdio>
#include<string.h>
#include<stack>
using namespace std;
int N;
const int maxn=50;
int preorder[maxn];
int inorder[maxn];
struct Node{
int val;
Node *lchild,*rchild;
};
Node *create(int preL,int preR,int inL,int inR){
if(preL > preR){
return NULL;
}
Node *root = new Node;
root->val = preorder[preL];
//找到先序中的根在中序遍历中的位置
int k;
for(k=inL;k<=inR;k++)
if(inorder[k]==preorder[preL])
break;
int numLeft= k - inL;//左子树结点个数
root->lchild=create(preL+1,preL+numLeft,inL,k-1);
root->rchild=create(preL+numLeft+1,preR,k+1,inR);
return root;
}
int cnt=0;
void postVisit(Node *p){
if(p!=NULL){
postVisit(p->lchild);
postVisit(p->rchild);
printf("%d",p->val);
cnt++;
if(cnt<N) printf(" ");
}
}
//给出先序和中序遍历 重建二叉树
int main(){
stack<int>st;
char str[5];
int pre_index=0,in_index=0;
scanf("%d",&N);
for(int i=0;i<2*N;i++){
scanf("%s",str);
if(strcmp("Push",str)==0){
int x;
scanf("%d",&x);
st.push(x);
preorder[pre_index++]=x;
}
else{
inorder[in_index++]=st.top();
st.pop();
}
}
/*
for(int i=0;i<N;i++)
{
printf("%d %d\n",preorder[i],inorder[i]);
}*/
Node *root=create(0,N-1,0,N-1);
postVisit(root);
return 0;
}
ps
这题我一个写法错了好久T.T
int k;
for(int k=inL;k<=inR;k++)
if(inorder[k]==preorder[preL])
break;
就是这种写法，里面的k是正确的值，外层的k一直是0，然后卡死了…
错误原因 循环内部的局部变量在for循环结束内存会释放
