我正在使用我的代码发布新问题。我试图计算二叉搜索树的非叶节点。我正在创建非叶方法,然后我试图在测试类中调用它。有人能帮我吗?这是代码:
public class BST {
private Node<Integer> root;
public BST() {
root = null;
}
public boolean insert(Integer i) {
Node<Integer> parent = root, child = root;
boolean gLeft = false;
while (child != null && i.compareTo(child.data) != 0) {
parent = child;
if (i.compareTo(child.data) < 0) {
child = child.left;
gLeft = true;
} else {
child = child.right;
gLeft = false;
}
}
if (child != null)
return false;
else {
Node<Integer> leaf = new Node<Integer>(i);
if (parent == null)
root = leaf;
else if (gLeft)
parent.left = leaf;
else
parent.right = leaf;
return true;
}
}
public boolean find(Integer i) {
Node<Integer> n = root;
boolean found = false;
while (n != null && !found) {
int comp = i.compareTo(n.data);
if (comp == 0)
found = true;
else if (comp < 0)
n = n.left;
else
n = n.right;
}
return found;
}
public int nonleaf() {
int count = 0;
Node<Integer> parent = root;
if (parent == null)
return 0;
if (parent.left == null && parent.right == null)
return 1;
}
}
class Node<T> {
T data;
Node<T> left, right;
Node(T o) {
data = o;
left = right = null;
}
}
答案 0 :(得分:0)
如果您只对非叶节点的计数感兴趣,那么您可以遍历树一次并保持一个计数。无论何时遇到节点,使其具有左或右节点增量计数。
答案 1 :(得分:0)
您可以使用以下函数来计算二叉树的非叶节点的数量。
int countNonLeafNodes(Node root)
{
if (root == null || (root.left == null &&
root.right == null))
return 0;
return 1 + countNonLeafNodes(root.left) +
countNonLeafNodes(root.right);
}