Java 基于线性阵列的二叉树构造方法
注意：这是家庭作业，所以请不要马上给我答案，但指针会很有帮助 任务概述： （测试每种方法以确保其正常工作。）Java 基于线性阵列的二叉树构造方法,java,arrays,binary-tree,Java,Arrays,Binary Tree,注意：这是家庭作业，所以请不要马上给我答案，但指针会很有帮助 任务概述： （测试每种方法以确保其正常工作。） 生成按正确顺序添加100个元素的二进制搜索-将数字1到100按正确顺序添加到树中添加1、2、3等 生成一个包含100个元素的二进制搜索，按相反顺序添加-添加数字100、99、98等等 生成包含100个随机生成元素的二进制搜索 我在问题中提到线性数组的原因是因为根必须是数组中的第一个元素，然后将其余元素相加 我没有所有其他的方法来测试这棵树是否制作正确，我马上就要去工作了。然而，我在下面发
public BST(int[] numbers)
{
node root;
node currentParent;
node previousParent = null;
int i = 1; // skip the root, that will be handled manualy
root = new node(numbers[0]); // first element is the root
currentParent = root;
// handle right hand side of binary tree first
while (i > root.getValue())
{
while (i > currentParent.getValue())
{
node newNode = new node (numbers[i]);
currentParent.setRightChild(newNode);
previousParent = currentParent;
currentParent = newNode;
i++;
} // end inner while
if (previousParent.leftChild == null)
{
node newNode = new node(numbers[i]);
previousParent.leftChild = newNode;
currentParent = newNode;
i++;
} // end if branch
else
{
System.out.println("Hmm, something seems to be wrong!");
} // end else
} // end right side binary tree while
} // end integer array constructor
public class node
{
node leftChild; // left pointer
node rightChild; // right pointer
int value; // will be the int value inside the node
boolean isLeaf; // boolean for determing if a node is a left node or not
/*
Null constructor for the class. Sets the pointers to null
*/
public node ()
{
leftChild = null;
rightChild = null;
} // end null constructor
/*
Intialzing constructor that will take a single integer as input
*/
public node (int number)
{
leftChild = null;
rightChild = null;
value = number;
} // end intialzing constructor with int input
/*
The setValue method will take the int value inputted from the array into
a new node
*/
public void setValue(int number)
{
value = number;
} // end setValue()
/*
The getValue method will return to the user the value that is stored
inside a specific node.
*/
public int getValue()
{
return value;
} // end getValue()
/*
The setLeftChild method will set the pointers to the nodes left child
which will be a value that is equal or lower to the paret node
*/
public void setLeftChild (node leftChild)
{
this.leftChild = leftChild;
} // end setLeftChild()
/*
The setRightChild method will be the same as setLeftChild above but will
handle setting the nodes right child. The right child will be a value
that is greater than the parent node
*/
public void setRightChild(node rightChild)
{
this.rightChild = rightChild;
} // end setRightChild
/*
the getLeftChild method will return to the user/calling method what the
left child of a given node is.
*/
public node getLeftChild (node currentNode)
{
return leftChild;
} // end getLeftChild()
/*
the getRightChild method is exactly the same as the getLeftChil method
above, except it will return the right child instead
*/
public node getRightChild(node currentNode)
{
return rightChild;
} // end getRightChild()
/*
The setLeaf method will be in charge of manipulating a boolean value and
setting it to true if the node is a left or not
*/
public void setLeaf(boolean leaf)
{
if (leaf == true)
{
isLeaf = true;
} // end if
else
{
isLeaf = false;
} // end else
} // end setLeaf
/*
The getLeaft method will simply return a boolean value that indicates if
the given node is a left node or not
*/
public boolean getLeaf()
{
return isLeaf;
} // end getLeaf()
} // end node class
不，对不起，这是错的
最后一个提示：类应命名为
NodeNode节点粘贴代码吗？此外，作业听起来像是你需要做一些类似于tree.add（1）的事情；树。添加（2）
我添加了节点类，而不是让构造函数从int[]
@Captain Man生成树。虽然我确实需要一个insert方法，但我确实需要一个以int数组为参数的构造函数parameter@Jeffery哦，我明白了。这棵树需要“平衡”还是“自我平衡”？这是值得考虑的问题。由于必须有一个insert方法，我建议您只需为数组中的每个int
调用insert方法，这样您就不会重复代码。类似于'for（inti=0；inumbers[I]