[英]How to count non-leaf nodes in a binary search tree?
我正在用我的代碼發布新問題。 我正在嘗試計算二進制搜索樹的非葉節點。 我正在創建非葉子方法,然后嘗試在測試類中調用它。 有人能幫我嗎? 這是代碼:
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;
}
}
如果您只對非葉節點的計數感興趣,則可以遍歷樹一次並保持一個計數。 每當遇到一個具有左右節點增量計數的節點時。
您可以使用以下函數來計算二叉樹的非葉節點的數量。
int countNonLeafNodes(Node root)
{
if (root == null || (root.left == null &&
root.right == null))
return 0;
return 1 + countNonLeafNodes(root.left) +
countNonLeafNodes(root.right);
}
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