leetcode--Lowest Common Ancestor of a Binary Search Tree

最新推荐文章于 2025-12-06 07:43:37 发布

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本文介绍如何在二叉搜索树中找到两个指定节点的最低公共祖先。提供了两种实现方法，一种使用递归，另一种为迭代方式。通过比较节点值来定位目标节点。

题目：Lowest Common Ancestor of a Binary Search Tree

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

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

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

题目分析：在查询树，寻找最小共同祖先。查找树：有序的树，左小右大；最小共同祖先：最近的一个公共父节点。

One：本质是比较p、q和每次root值的大小，用递归的方法

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
         if (root == null || p == null || q == null) return null;
         if(root.val<Math.min(p.val,q.val)) return lowestCommonAncestor(root.right,p,q);
         if(root.val>Math.max(p.val,q.val)) return lowestCommonAncestor(root.left,p,q);
         return root;
    }
}

Two：不用递归的方法，普通的方法

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while (true) {
            if (root == null || root == p || root == q) {return root;}
            if (root.val < Math.min(p.val,q.val)) {root = root.right;}
            else if (root.val > Math.max(p.val,q.val)) {root = root.left;}
            else {return root;}
        }
    }