我编写了下面的代码来检查一棵树是否为二叉搜索树。请帮忙检查一下代码：
好的！代码现在已经被编辑过了。这个简单的解决方案是在下面的帖子中由某人提出的：
IsValidBST(root,-infinity,infinity);
bool IsValidBST(BinaryNode node, int MIN, int MAX)
{
if(node == null)
return true;
if(node.element > MIN
&& node.element < MAX
&& IsValidBST(node.left,MIN,node.element)
&& IsValidBST(node.right,node.element,MAX))
return true;
else
return false;
}
没错，另一个简单的解决方案是进行中序遍历。
public boolean isBst(BNode node)
{
return isBinarySearchTree(node , Integer.MIN_VALUE , Integer.MIN_VALUE);
}
内部实现：
public boolean isBinarySearchTree(BNode node , int min , int max)
{
if(node.data < min || node.data > max)
return false;
//Check this node!
//This algorithm doesn't recurse with null Arguments.
//When a null is found the method returns true;
//Look and you will find out.
/*
* Checking for Left SubTree
*/
boolean leftIsBst = false;
//If the Left Node Exists
if(node.left != null)
{
//and the Left Data are Smaller than the Node Data
if(node.left.data < node.data)
{
//Check if the subtree is Valid as well
leftIsBst = isBinarySearchTree(node.left , min , node.data);
}else
{
//Else if the Left data are Bigger return false;
leftIsBst = false;
}
}else //if the Left Node Doesn't Exist return true;
{
leftIsBst = true;
}
/*
* Checking for Right SubTree - Similar Logic
*/
boolean rightIsBst = false;
//If the Right Node Exists
if(node.right != null)
{
//and the Right Data are Bigger (or Equal) than the Node Data
if(node.right.data >= node.data)
{
//Check if the subtree is Valid as well
rightIsBst = isBinarySearchTree(node.right , node.data+1 , max);
}else
{
//Else if the Right data are Smaller return false;
rightIsBst = false;
}
}else //if the Right Node Doesn't Exist return true;
{
rightIsBst = true;
}
//if both are true then this means that subtrees are BST too
return (leftIsBst && rightIsBst);
}
/*
* A Class which is used when getting subTrees Values
*/
class TreeValues
{
BNode root; //Which node those values apply for
int Min;
int Max;
TreeValues(BNode _node , _min , _max)
{
root = _node;
Min = _min;
Max = _max;
}
}
并且：
/*
* Use this as your container to store Min and Max of the whole
*/
ArrayList<TreeValues> myValues = new ArrayList<TreeValues>;
现在这是一种方法，用于查找给定节点的最小值和最大值:
/*
* Method Used to get Values for one Subtree
* Returns a TreeValues Object containing that (sub-)trees values
*/
public TreeValues GetSubTreeValues(BNode node)
{
//Keep information on the data of the Subtree's Startnode
//We gonna need it later
BNode SubtreeRoot = node;
//The Min value of a BST Tree exists in the leftmost child
//and the Max in the RightMost child
int MinValue = 0;
//If there is not a Left Child
if(node.left == null)
{
//The Min Value is this node's data
MinValue = node.data;
}else
{
//Get me the Leftmost Child
while(node.left != null)
{
node = node.left;
}
MinValue = node.data;
}
//Reset the node to original value
node = SubtreeRoot; //Edit - fix
//Similarly for the Right Child.
if(node.right == null)
{
MaxValue = node.data;
}else
{
int MaxValue = 0;
//Similarly
while(node.right != null)
{
node = node.right;
}
MaxValue = node.data;
}
//Return the info.
return new TreeValues(SubtreeRoot , MinValue , MaxValue);
}
public void GetTreeValues(BNode node)
{
//Add this node to the Container with Tree Data
myValues.add(GetSubTreeValues(node));
//Get Left Child Values, if it exists ...
if(node.left != null)
GetTreeValues(node.left);
//Similarly.
if(node.right != null)
GetTreeValues(node.right);
//Nothing is returned, we put everything to the myValues container
return;
}
if(isBinarySearchTree(root))
GetTreeValues(root);
//else ... Do Something
这几乎是Java。只需要进行一些修改和修复就可以运行。找一本好的面向对象编程书籍，它会对你有所帮助。请注意，这个解决方案可以拆分成更多的方法。
boolean bst = isBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
public boolean isBST(Node root, int min, int max) {
if(root == null)
return true;
return (root.data > min &&
root.data < max &&
isBST(root.left, min, root.data) &&
isBST(root.right, root.data, max));
}
更新：我刚刚看到此解决方案之前已经被建议过了。抱歉，也许仍然有人会发现我的版本有用。
这里提供一个解决方案，使用中序遍历来检查BST属性。 在提供解决方案之前，请注意我使用的BST定义不允许有重复项。这意味着BST中的每个值都是独特的（只是为了简单起见）。
递归中序打印的代码：
void printInorder(Node root) {
if(root == null) { // base case
return;
}
printInorder(root.left); // go left
System.out.print(root.data + " "); // process (print) data
printInorder(root.right); // go right
}
5
3 7
1 2 6 9
需要打印的内容：
如果按顺序打印，应该是：1 2 3 5 6 7 9
5
3 7
1 8 6 9
存在违规情况。节点3的右子节点是8，如果3是根节点，那么这样做是可以的。但在二叉搜索树中，8将成为9的左子节点而不是3的右子节点。因此，该树不是二叉搜索树。
因此，代码应按照以下思路编写：
/* wrapper that keeps track of the previous value */
class PrevWrapper {
int data = Integer.MIN_VALUE;
}
boolean isBST(Node root, PrevWrapper prev) {
/* base case: we reached null*/
if (root == null) {
return true;
}
if(!isBST(root.left, prev)) {
return false;
}
/* If previous in-order node's data is larger than
* current node's data, BST property is violated */
if (prev.data > root.data) {
return false;
}
/* set the previous in-order data to the current node's data*/
prev.data = root.data;
return isBST(root.right, prev);
}
boolean isBST(Node root) {
return isBST(root, new PrevWrapper());
}
boolean isBST(TreeNode root, int max, int min) {
if (root == null) {
return true;
} else if (root.val >= max || root.val <= min) {
return false;
} else {
return isBST(root.left, root.val, min) && isBST(root.right, max, root.val);
}
}
#include <limits>
int min = numeric_limits<int>::min();
int max = numeric_limits<int>::max();
The calling function will pass the above min and max values initially to isBst(...)
bool isBst(node* root, int min, int max)
{
//base case
if(root == NULL)
return true;
if(root->val <= max && root->val >= min)
{
bool b1 = isBst(root->left, min, root->val);
bool b2 = isBst(root->right, root->val, max);
if(!b1 || !b2)
return false;
return true;
}
return false;
}
import java.util.Stack;
final int MIN_INT = Integer.MIN_VALUE;
final int MAX_INT = Integer.MAX_VALUE;
public class NodeBounds {
BinaryTreeNode node;
int lowerBound;
int upperBound;
public NodeBounds(BinaryTreeNode node, int lowerBound, int upperBound) {
this.node = node;
this.lowerBound = lowerBound;
this.upperBound = upperBound;
}
}
public boolean bstChecker(BinaryTreeNode root) {
// start at the root, with an arbitrarily low lower bound
// and an arbitrarily high upper bound
Stack<NodeBounds> stack = new Stack<NodeBounds>();
stack.push(new NodeBounds(root, MIN_INT, MAX_INT));
// depth-first traversal
while (!stack.empty()) {
NodeBounds nb = stack.pop();
BinaryTreeNode node = nb.node;
int lowerBound = nb.lowerBound;
int upperBound = nb.upperBound;
// if this node is invalid, we return false right away
if ((node.value < lowerBound) || (node.value > upperBound)) {
return false;
}
if (node.left != null) {
// this node must be less than the current node
stack.push(new NodeBounds(node.left, lowerBound, node.value));
}
if (node.right != null) {
// this node must be greater than the current node
stack.push(new NodeBounds(node.right, node.value, upperBound));
}
}
// if none of the nodes were invalid, return true
// (at this point we have checked all nodes)
return true;
}
public void inorder()
{
min=min(root);
//System.out.println(min);
inorder(root);
}
private int min(BSTNode r)
{
while (r.left != null)
{
r=r.left;
}
return r.data;
}
private void inorder(BSTNode r)
{
if (r != null)
{
inorder(r.getLeft());
System.out.println(r.getData());
if(min<=r.getData())
{
System.out.println(min+"<="+r.getData());
min=r.getData();
}
else
System.out.println("nananan");
//System.out.print(r.getData() +" ");
inorder(r.getRight());
return;
}
}
将INTEGER.MIN和INTEGER.MAX作为空树的返回值并不是很合理。也许可以使用Integer类型，然后返回null。
{}）来格式化您的代码以便更易于阅读。最后，您的代码有什么问题？我们到底在“检查”什么，您是否遇到错误？ - LittleBobbyTables - Au Revoir