英文:
how to use only 1 return in this recursive method
问题
// 有多少个节点恰好有一个非空子节点
public int stickCt() {
return stickCt(root);
}
private int stickCt(StringNode t) {
int count = 0;
if (t == null)
return 0;
else if ((t.getLeft() == null && t.getRight() != null) || (t.getLeft() != null && t.getRight() == null))
count++;
count = count + stickCt(t.getLeft()) + stickCt(t.getRight());
return count;
}
英文:
I'm finishing up the easy portion project and my professor specified that he only wants methods with at maximum 1 return. I can't seem to figure out how to make this work correctly with only 1 return though. For context, I am finding how many nodes in a tree have exactly 1 non-null children.
// How many nodes have exactly 1 non-null children
public int stickCt() {
return stickCt(root);
}
private int stickCt(StringNode t) {
int count = 0;
if (t == null)
return 0;
else if ((t.getLeft() == null && t.getRight() != null) || (t.getLeft() != null & t.getRight() == null))
count++;
count = count + stickCt(t.getLeft()) + stickCt(t.getRight());
return count;
}
答案1
得分: 1
以下是翻译好的部分：
private int stickCt(StringNode t) {
int count = 0;
if (t != null) {
if ((t.getLeft() == null && t.getRight() != null) || (t.getLeft() != null && t.getRight() == null))
count++;
count += stickCt(t.getLeft()) + stickCt(t.getRight());
}
return count;
}
然而，在这种情况下，我发现您使用两个返回语句的表述更容易理解。
英文:
Something like this:
private int stickCt(StringNode t) {
int count = 0;
if (t != null) {
if ((t.getLeft() == null && t.getRight() != null) || (t.getLeft() != null & t.getRight() == null))
count++;
count += stickCt(t.getLeft()) + stickCt(t.getRight());
}
return count;
}
However, in this case I find your formulation with the 2 returns easier to understand.
答案2
得分: 0
使用递归算法的替代方法是使用广度优先搜索（BFS）或深度优先搜索（DFS）算法。
以下是一个BFS方法的示例：
public int stickCt(StringNode root){
int count = 0;
Queue<StringNode> queue = new LinkedList<StringNode>();
queue.add(root);
while (!queue.isEmpty()){
StringNode tempNode = queue.poll();
if (tempNode != null && (tempNode.getLeft() == null && tempNode.getRight() != null) || (tempNode.getLeft() != null && tempNode.getRight() == null)){
count++;
}
/*将左子节点入队*/
if (tempNode.getLeft() != null) {
queue.add(tempNode.getLeft());
}
/*将右子节点入队*/
if (tempNode.getRight() != null) {
queue.add(tempNode.getRight());
}
}
return count;
}
注意：由于您请求只返回翻译的部分，我已经将代码部分翻译为中文。如果您有任何其他问题或需要进一步帮助，请随时提问。
英文:
Instead using recursive algorithm you can use BFS of DFS algorithm.
Here is an example or BSF approach
public int stickCt(StringNode root){
int count = 0;
Queue<StringNode > queue = new LinkedList<StringNode >();
queue.add(root);
while (!queue.isEmpty()){
StringNode tempNode = queue.poll();
if (tempNode != null && (tempNode.getLeft() == null && tempNode.getRight() != null) || (tempNode.getLeft() != null && tempNode.getRight() == null)){
count++;
}
/*Enqueue left child */
if (tempNode.getLeft()!= null) {
queue.add(tempNode.getLeft());
}
/*Enqueue right child */
if (tempNode.getRight()!= null) {
queue.add(tempNode.getRight());
}
}
return count;
}
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