Algorithm-nearest common ancestor of binary tree

1. The nearest common ancestor of the binary tree (2 nodes)

236. The nearest common ancestor of the binary tree

给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉树:  root = [3,5,1,6,2,0,8,null,null,7,4]

示例 1:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
输出: 3
解释: 节点 5 和节点 1 的最近公共祖先是节点 3。
示例 2:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
输出: 5
解释: 节点 5 和节点 4 的最近公共祖先是节点 5。因为根据定义最近公共祖先节点可以为节点本身。
 

说明:

所有节点的值都是唯一的。
p、q 为不同节点且均存在于给定的二叉树中。

The title description is very clear. The title has given us the conditions, saying that the node must exist in the binary tree given to us, then we can implement our algorithm on this basis.

We can imagine that there are two situations:
1. Two nodes have a parent node that is not equal to themselves
2. One of the two nodes is the parent of the other node

Based on the above conditions, we can get the following conclusions

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    
    
        if(root==null||p==root||q==root){
    
    
            return root;
        }
        //left和right一定是p或q或者null,一定不可能同时为null,除非pq都为null
        //如果都不为null，那么说明满足我们上面提到的情况1
        //如果有一个为null，那么满足情况2
        TreeNode left=lowestCommonAncestor(root.left,p,q);
        TreeNode right=lowestCommonAncestor(root.right,p,q);
        if(left!=null&&right!=null){
    
    
            return root;
        }
        return left!=null?left:right;
    }

See the notes for details, not much explanation here

2. The nearest common ancestor of the binary tree (3 nodes)

The three-node binary tree common ancestor problem is exactly the same as the two-node or even n-node logic, which is very simple and will not be repeated.

    public TreeNode threeNodesHead(TreeNode root,TreeNode n1, TreeNode n2, TreeNode n3){
    
    
        return originNode(root,n1,n2,n3);
    }

    public TreeNode originNode(TreeNode root,TreeNode n1,TreeNode n2,TreeNode n3){
    
    
        if(root==null){
    
    
            return null;
        }
        if(n1==root||n2==root||n3==root){
    
    
            return root;
        }
        TreeNode left=originNode(root.left,n1,n2,n3);
        TreeNode right=originNode(root.right,n1,n2,n3);
        if(left!=null&&right!=null){
    
    
            return root;
        }
        if(left!=null){
    
    
            return left;
        }
        if(right!=null){
    
    
            return right;
        }
        return null;

    }