LeetCode --- 235. Lowest Common Ancestor of a Binary Search Tree

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#235. Lowest Common Ancestor of a Bi #Lowest Common Ancestor of a Binary

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本文探讨了在二叉搜索树中寻找两个节点的最低公共祖先(LCA)的算法，提供了递归与非递归两种解决方案，并详细解析了代码实现，强调了节点值的唯一性和目标节点的存在。

235. Lowest Common Ancestor of a Binary Search Tree

Difficulty: Easy

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

According to the : “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.

Solution

思路1

递归遍历，时间复杂度O(n)。如果root大于p和q的值，说明p和q在左子树，反之在右子树。代码如下：
Language: C++

/**
 * 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 root;
        if (root->val < p->val && root->val < q->val)
            return lowestCommonAncestor(root->right, p, q);
        else if (root->val > p->val && root->val > q->val)
            return lowestCommonAncestor(root->left, p, q);
        else
            return root;
    }
};

思路2

非递归，代码如下；

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