Java中的二進制搜索樹可視化

Question

嗨，我目前正在項目的測試階段（算法可視化工具）。 我的BST的刪除方法遇到問題。

 public boolean delete(String key) {
boolean deleted = true;
boolean finished=false;
BNode current = root;
BNode prev = null;
while (!finished) {
  if (key.compareTo(current.key) > 0) {
    prev = current;
    current = current.right;
    this.repaint();
  }
  else if (key.compareTo(current.key) < 0) {
    prev = current;
    current = current.left;
    this.repaint();
  }
  else if (key.compareTo(current.key) == 0) {
      finished=true;
      this.repaint();
  }

}

if (check(current) == 0) {
    if(current==root)
    {
        root=null;
        xPos=400;
        yPos=60;
        this.repaint();
    }
    else
    {
        if (current.key.compareTo(prev.key) > 0) {
            prev.right = null;
            this.repaint();
        }
        else if(current.key.compareTo(prev.key) < 0) {
            prev.left = null;
            this.repaint();
        }
    }

}
else if (check(current) == 1) {
    if(current==root)
    {
        prev=current;
        if (current.left != null) {
            current=current.left;
            prev.key=current.key;
            prev.left = current.left;
            this.repaint();
        }
        else {
            current=current.right;
            prev.key=current.key;
            prev.right = current.right;
            this.repaint();
        }
    }
    else
    {

    if (current.key.compareTo(prev.key) > 0) {
    if (current.left != null) {
      prev.right = current.left;
      this.repaint();
    }
    else {
      prev.right = current.right;
      this.repaint();
    }
  }
  else if(current.key.compareTo(prev.key) < 0) {
    if (current.left != null) {
      prev.left = current.left;
      this.repaint();
    }
    else {
      prev.left = current.right;
      this.repaint();
    }
  }
    }
}
else if (check(current) == 2) {
  BNode temp = inord(current);
  if(current==root)
  {
      root.key=temp.key;
      this.repaint();
  }
  else
  {

      if (current.key.compareTo(prev.key) > 0) {
      prev.right.key = temp.key;
      this.repaint();
    }
    else {
      prev.left.key = temp.key;
      this.repaint(0);
    }
    }
}

return deleted;}

BST類本身的代碼更長。 一切工作正常，除了當我嘗試刪除不帶子節點的節點時，當我使用9和10作為輸入（嘗試對del 10）或5和12（嘗試對del 12）時，出現空指針異常，但是從不如果我使用例如4和8（嘗試刪除8）或9、6和5，我認為問題出在compareTo。

int check(BNode a) {
int ret;
if ( (a.left != null) && (a.right != null)) {
  ret = 2;
}
else if ( (a.left == null) && (a.right == null)) {
  ret = 0;
}
else {
  ret = 1;
}
return ret;}

我真的需要幫助。如有需要，我可以發布整個課程。謝謝！

Answer 1

只是一些注意事項：

1. 如果傳遞null進行檢查，則將在那里獲得NPE。
2. if( check(current) == 0)等->您應該檢查一次，然后執行if（甚至切換）

示例2：

 int result = check(current);
 switch(result) {
  case 0:
    //do whatever is appropriate
    break;
  case 1:
    //do whatever is appropriate
    break;
  case 2:
    //do whatever is appropriate
    break;
  default:
    //should never happen, either leave it or throw an exception if it ever happens
}

編輯：//實際上，忘記了此編輯，只是看到了這種情況不應該發生，但是它仍然不是一個好的樣式

您的代碼中也包含以下內容：

if (current.left != null) {
    current=current.left;
    prev.key=current.key;
    prev.left = current.left;
    this.repaint();
}
else {
    current=current.right; //this might be null
 ...
}

如果current.left為null並且current.right為null，則current之后將為null。