查找树搜索中的最低命令祖先失败（不是算法问题，而是Java问题）

Question

非常简单的树问题，结果是null（应该是1，因为1是2&3的父节点），找不到原因。该方法本身已通过Leetcode在线判断。

以下是问题的链接：

public class LowestCommonAncestor2 {

    public static TreeNode mockTree() {
        TreeNode node1 = new TreeNode(1);
        TreeNode node2 = new TreeNode(2);
        TreeNode node3 = new TreeNode(3);
        node1.left = node2;
        node1.right = node3;
        return node1;
    }

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode t1, TreeNode t2) {
           if (root == null || root == t1 || root == t2) {
               return root;
           }

           TreeNode left = lowestCommonAncestor(root.left, t1, t2);
           TreeNode right = lowestCommonAncestor(root.right, t1, t2);

           if (left != null && right != null) 
               return root;
           if (left != null) 
               return left;
           if (right != null) 
               return right;
           return null;
    }

    public static void main (String [] args) {
        LowestCommonAncestor2 test = new LowestCommonAncestor2();
        TreeNode root = mockTree();
        TreeNode target1 = new TreeNode(2);
        TreeNode target2 = new TreeNode(3);
        TreeNode result = test.lowestCommonAncestor(root, target1, target2);
        System.out.println(result);
    } 
}

和TreeNode的定义如下：

public class TreeNode {
     public int val;
     public TreeNode left, right;
     public TreeNode(int val) {
         this.val = val;
         this.left = this.right = null;
     }
}

Answer 1

发生错误是因为这行代码

if (root == null || root == t1 || root == t2) 

在执行此方法时，请仔细考虑您要比较的是什么

TreeNode result = test.lowestCommonAncestor(root, target1, target2);

当您调用LCA方法时，第一个if语句中的任何条件都不会计算为true，因为没有一个语句是true，这是正确的。但是，现在让我们来讨论第一个递归调用，

TreeNode left = lowestCommonAncestor(root.left, t1, t2);
TreeNode right = lowestCommonAncestor(root.right, t1, t2);

在TreeNode类中创建一个equals方法，该方法返回一个布尔值并接受另一个TreeNode对象作为参数。此外，请确保此方法是实例方法而不是静态方法，以便在代码中执行此方法时如下所示，

public boolean equals(TreeNode node){ //method address
    //rest of the method has to be implemented yourself
    //ask yourself, how are two treenodes considered equal in your situation?
}

最后，将root==t1替换为

root.equals(t1)

和根==t2

root.equals(t2)