235. Lowest Common Ancestor of a Binary Search Tree

最新推荐文章于 2025-08-17 23:50:11 发布

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最新推荐文章于 2025-08-17 23:50:11 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了一种寻找二叉搜索树中两给定节点的最低公共祖先的方法。利用二叉搜索树的特性，即左节点小于等于父节点而右节点大于父节点，通过递归地比较当前节点与目标节点的值来定位最低公共祖先。

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leetcode

题目描述

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.

解题思路

此题要清楚这是一颗二叉搜索树，二叉搜索树的特点是左边的节点小于等于父节点，右边的节点大于父节点。所以两个节点的最小公共祖先的大小应该是搜索到的第一个介于要查找的两个节点之间的值。
1. 如果找到的节点比两个要找的节点都大的话，那么就搜索该树的左子树。
2. 如果找到的节点比两个要找的节点都小的话，那么就搜索该树的右子树。

代码

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     * int val;
     * TreeNode *left;
     * TreeNode *right;
     * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root) return nullptr;
        TreeNode* answer = nullptr;
        CommonAncestor(root, p, q, answer);
        return answer;
    }
    void CommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q,  TreeNode*& ans)
    {
        if(!root) return;
        if(root->val >= p && root->val <= q || root->val >= q && root->val <= p)
        {
            ans = root;
            return;
        }
        if(root->val > p && root->val >q)
            CommonAncestor(root->left, p, q, ans);
        if(root->val < p && root->val < q)
            CommonAncestor(root->right, p, q, ans);
    }
    };