1043. Is It a Binary Search Tree (25)

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line "YES" if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or "NO" if not. Then if the answer is "YES", print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

78 6 5 7 10 8 11

Sample Output 1:

YES5 7 6 8 11 10 8

Sample Input 2:

78 10 11 8 6 7 5

Sample Output 2:

YES11 8 10 7 5 6 8

Sample Input 3:

78 6 8 5 10 9 11

Sample Output 3:

NO

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给出一棵树的前序遍历，判断这棵数是否是二分搜索树。可以发现二分搜索树前序遍历的规律，第一个点为根节点，往后比它小的点是它的左子树的点，再往后的全部是它右子树的点，应该全部大于等于它，由这个规律可以递归判断是否是二叉搜索树。只要有某个子树不符合规律，就不是二叉搜索树。这里还要判断这棵树的镜像是不是二叉搜索树，这里只需要上面的比较反过来（就是判断小于变成判断大于等于）就行了，这里将函数作为参数，以便重用同一个函数来进行判断。最后递归实现后续遍历这棵树。

代码：

#include <iostream>#include <cstring>#include <vector>#include <cstdlib>#include <cstdio>using namespace std;struct node{int val;node *left;node *right;};vector<int>a;bool isBST=true;bool gt(const int &a,const int &b){return a>=b;}bool lt(const int &a,const int &b){return a<b;}node* build(int l,int r,bool cmp(const int&,const int&)){if(!isBST) return NULL;if(l>r) return NULL;node *root=new node();root->val=a[l];if(l==r) return root;int mid;for(mid=l+1;mid<=r;mid++){if(!cmp(a[mid],a[l])) {break;}}root->left=build(l+1,mid-1,cmp);int j;for(j=mid;j<=r;j++){if(cmp(a[j],a[l])){isBST=false;return NULL;}}root->right=build(mid,r,cmp);return root;}bool flg=true;void postorder(node *root){if(root){postorder(root->left);postorder(root->right);if(flg){cout<<root->val;flg=false;}else{cout<<" "<<root->val;}}}int main(){int n;cin>>n;a.resize(n);for(int i=0;i<n;i++){cin>>a[i];}if(n==1){cout<<"YES"<<endl;cout<<a[0]<<endl;return 0;}node *root=build(0,n-1,lt);if(isBST){cout<<"YES"<<endl;postorder(root);}else{isBST=true;root=build(0,n-1,gt);if(isBST){cout<<"YES"<<endl;postorder(root);}else cout<<"NO"<<endl;}}