到目前为止,我已经找到了可以添加到二进制搜索树中的算法,但是在将其转换为代码方面遇到了一些困难。算法如下:
public void add(int v) {
Create a new node n to hold value v.
If tree is empty
Set root to n.
Else
Create temporary node reference m, initialized to root.
Loop looking for a gap (a null left or right pointer) that is on the
correct side of m for v
If v < m.value, look at the left pointer
If v >= m.value, look at the right pointer
If pointer on correct side is not null, set m to that child node and
continue looking
m = m.left or m = m.right
The search for insertion position stops when node m has a null pointer on
the correct side.
Insert the new node n at that position
m.left = n or m.right = n
}
到目前为止,我有:
public void add(int v) {
Node n = new Node(v);
if(root==null)
root = n;
else {
Node m = root;
while(...) {
if(...)
m = m.left;
else
m = m.right;
}
if(...)
m.left = m;
else
m.right = n;
}
}
我相信大多数方法都是正确的,但是我不知道在标记为“ ...”的地方需要做什么。
最佳答案
首先,二进制搜索树不应有任何重复的值,这是您尚未在代码中实现的重要要求。我最近在学习Java数据结构时实现了二进制搜索树。这是我写的代码:
public class OrderedBinaryTree
{
private int _elementsPresent = 0;
private Node _root = null;
private int [] _values = null;
private class Node
{
Node _left = null;
Node _right = null;
Node _parent = null;
int _value = 0;
public Node(int value,Node parent)
{
_value = value;
_parent = parent;
}
}
public void put(int value)
{
boolean valueInserted = false;
Node temp = _root;
while(!valueInserted)
{
if(_root == null)
{
_root = new Node(value,null);
break;
}
else if(value == temp._value)
{
System.out.println("the entered value is already present");
return;
}
else if(value<=temp._value)
{
if(temp._left == null)
{
temp._left = new Node(value,temp);
break;
}
else
{
temp = temp._left;
}
}
else
{
if(temp._right == null)
{
temp._right = new Node(value,temp);
break;
}
else
{
temp = temp._right;
}
}
}
_elementsPresent++;
}