Java 简单的二叉树检查等于
Java Simple Binary Tree check equals
我不明白为什么这个 equals
函数不起作用:
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
public boolean equals(TreeNode other) {
if (this == null && other == null) {
return true;
} else if (this == null || other == null) {
return false;
} else {
return this.val == other.val
&& this.left.equals(other.left)
&& this.right.equals(other.right);
}
}
}
看来主要问题是我无法比较 null TreeNode,但在设置中我已经指出了如何处理 null?
TreeNode a = new TreeNode(5);
TreeNode b = new TreeNode(5);
System.out.println(a.equals(b)); // >>> NullPointerException
上面的比较从非空开始,但最终当分支到空左或空右时它会命中空。一种工作方式是将上面相同的方法提取到静态方法中:
public static boolean isEqual(TreeNode self, TreeNode other) {
if (self == null && other == null) {
return true;
} else if (self == null || other == null) {
return false;
} else {
return self.val == other.val
&& isEqual(self.left, other.left)
&& isEqual(self.right, other.right);
}
}
这会很好用:
TreeNode a = new TreeNode(5);
TreeNode b = new TreeNode(5);
System.out.println(TreeNode.isEqual(a, b)); // >>> true
下面也可以,避免this.left/right
变成null
,看起来很傻,但确实是java
public boolean equals(TreeNode other) {
if (this.left == null && other.left == null && this.right == null && other.right == null) {
return this.val == other.val;
} else if (this.left != null && other.left != null && this.right != null && other.right != null) {
return this.val == other.val && this.left.equals(other.left) && this.right.equals(other.right);
} else if (this.left != null && other.left != null && this.right == null && other.right == null) {
return this.val == other.val && this.left.equals(other.left);
} else if (this.left == null && other.left == null && this.right != null && other.right != null) {
return this.val == other.val && this.right.equals(other.right);
} else {
return false;
}
}
a 为空。因此,调用“equals”将立即引发 NullPointerException,而不会调用 equals 方法。
两个可能让您感到困惑的问题:
- 在第一个实现中,检查“this==null”是多余的(在此上下文中 this 不能为 null)。
- “this.left”和“this.right”空检查也是多余的,因为它们是原语,永远不能为空。
我认为应该是这样的:
public boolean equals(TreeNode other) {
if (other == null) {
return false;
}
// Different Values
if (this.val != other.val) {
return false;
}
if (this.left == null && other.left != null) {
return false;
}
if (this.left != null && !this.left.equals(other.left)) {
return false;
}
if (this.right == null && other.right != null) {
return false;
}
if (this.right != null && !this.right.equals(other.right)) {
return false;
}
return true;
}
我不明白为什么这个 equals
函数不起作用:
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
public boolean equals(TreeNode other) {
if (this == null && other == null) {
return true;
} else if (this == null || other == null) {
return false;
} else {
return this.val == other.val
&& this.left.equals(other.left)
&& this.right.equals(other.right);
}
}
}
看来主要问题是我无法比较 null TreeNode,但在设置中我已经指出了如何处理 null?
TreeNode a = new TreeNode(5);
TreeNode b = new TreeNode(5);
System.out.println(a.equals(b)); // >>> NullPointerException
上面的比较从非空开始,但最终当分支到空左或空右时它会命中空。一种工作方式是将上面相同的方法提取到静态方法中:
public static boolean isEqual(TreeNode self, TreeNode other) {
if (self == null && other == null) {
return true;
} else if (self == null || other == null) {
return false;
} else {
return self.val == other.val
&& isEqual(self.left, other.left)
&& isEqual(self.right, other.right);
}
}
这会很好用:
TreeNode a = new TreeNode(5);
TreeNode b = new TreeNode(5);
System.out.println(TreeNode.isEqual(a, b)); // >>> true
下面也可以,避免this.left/right
变成null
,看起来很傻,但确实是java
public boolean equals(TreeNode other) {
if (this.left == null && other.left == null && this.right == null && other.right == null) {
return this.val == other.val;
} else if (this.left != null && other.left != null && this.right != null && other.right != null) {
return this.val == other.val && this.left.equals(other.left) && this.right.equals(other.right);
} else if (this.left != null && other.left != null && this.right == null && other.right == null) {
return this.val == other.val && this.left.equals(other.left);
} else if (this.left == null && other.left == null && this.right != null && other.right != null) {
return this.val == other.val && this.right.equals(other.right);
} else {
return false;
}
}
a 为空。因此,调用“equals”将立即引发 NullPointerException,而不会调用 equals 方法。
两个可能让您感到困惑的问题:
- 在第一个实现中,检查“this==null”是多余的(在此上下文中 this 不能为 null)。
- “this.left”和“this.right”空检查也是多余的,因为它们是原语,永远不能为空。
我认为应该是这样的:
public boolean equals(TreeNode other) {
if (other == null) {
return false;
}
// Different Values
if (this.val != other.val) {
return false;
}
if (this.left == null && other.left != null) {
return false;
}
if (this.left != null && !this.left.equals(other.left)) {
return false;
}
if (this.right == null && other.right != null) {
return false;
}
if (this.right != null && !this.right.equals(other.right)) {
return false;
}
return true;
}