235. Lowest Common Ancestor of a Binary Search Tree

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最新推荐文章于 2025-09-13 20:19:55 发布

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本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/89292178

本文探讨了在二叉搜索树中寻找两个指定节点的最低公共祖先(LCA)问题，提供了递归和迭代两种解决方案，并分析了其时间与空间复杂度。通过实例说明了不同情况下的查找方向和条件。

235. Lowest Common Ancestor of a Binary Search Tree

方法1: recursion
- Complexity
- 易错点：
方法2: iterative

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 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 binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, 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 BST.

方法1: recursion

思路：

如果一个在左一个在右，当前节点必为lca。否则递归向左查找，或递归向右查找。

Complexity

Time complexity: O(h)
Space complexity: O(h)

易错点：

向左向右查找的方向及条件。

class Solution2 {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root -> val == p -> val || root -> val == q -> val) return root;
        if (root -> val > p -> val && root -> val > q -> val) {
            return lowestCommonAncestor(root -> left, p, q);
        }
        else if (root -> val < p -> val && root -> val < q -> val) {
            return lowestCommonAncestor(root -> right, p, q);
        }
        else {
            return root;
        }
    }
};

方法2: iterative

思路：

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        while (true) {
            if (root -> val < p -> val && root -> val < q -> val) {
                root = root -> right;
            }
            else if (root -> val > p -> val && root -> val > q -> val) {
                root = root -> left;
            }
            else return root;
        }
        return root;
    }
};