leetcode 236. 二叉树的最近公共祖先(java)

最新推荐文章于 2023-03-05 17:19:48 发布

翻译最新推荐文章于 2023-03-05 17:19:48 发布 · 768 阅读

leetcode二刷专栏收录该内容

17 篇文章

订阅专栏

本文探讨了在给定二叉树中寻找两个指定节点的最低公共祖先（LCA）的问题，提供了两种不同的实现方法，并通过具体示例说明了算法的工作原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

All of the nodes' values will be unique.
p and q are different and both values will exist in the binary tree.

代码：

一、我自己写的代码，看起来有点罗嗦。

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null)
            return null;
        TreeNode left=lowestCommonAncestor(root.left,p,q);
        if(left!=null)
            return left;
        TreeNode right=lowestCommonAncestor(root.right,p,q);
        if(right!=null)
            return right;
        if(includeNode(root,p)&&includeNode(root,q))
            return root;
        return null;
    }
    public boolean includeNode(TreeNode root, TreeNode p){//判断root位根的树是否包含节点p
        if(root!=null){
            if(root==p||includeNode(root.left,p)||includeNode(root.right,p)){
                System.out.println(root.val+":"+p.val);
                return true;
            }
        }
       return false;     
    }
}

二、别人的代码

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null||root==p||root==q){
            return root;
        }
        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;
    }
}