236. Lowest Common Ancestor of a Binary Tree

最新推荐文章于 2024-02-06 06:38:28 发布

N.C_IPOC_BUPT

最新推荐文章于 2024-02-06 06:38:28 发布

阅读量158

点赞数

CC 4.0 BY-SA版权

分类专栏： Tree

本文链接：https://blog.youkuaiyun.com/IPOC_BUPT/article/details/88855798

Tree 专栏收录该内容

84 篇文章

订阅专栏

本文介绍了解决二叉树中寻找两个节点的最低公共祖先（LCA）问题的两种方法。方法一采用迭代方式，通过查找指定节点来确定公共祖先；方法二则使用递归策略，效率更高。这两种方案均提供了高效的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述

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]

在这里插入图片描述

题目链接

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

方法思路

Approach1：

class Solution {
    //Runtime: 5 ms, faster than 100.00%
    //Memory Usage: 33 MB, less than 96.87% 
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(find(p, q.val)) return p;
        if(find(q, p.val)) return q;
        while(root != null){
            if(find(root.left, p.val) && find(root.left, q.val)){
                root = root.left;
                continue;
            }
            if(find(root.right, p.val) && find(root.right, q.val))
                root = root.right;
            else
                return root;
        }
        return null;
    }
    public boolean find(TreeNode root, int val){
        if(root == null) return false;
        if(root.val == val) return true;
        return find(root.left, val) || find(root.right, val);
    }
}

Approach2:recursive

public class Solution {
    //Runtime: 6 ms, faster than 75.86%
    //Memory Usage: 33.6 MB, less than 83.96% 
    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;
    }
}