leetcode235. Lowest Common Ancestor of a Binary Search Tree

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 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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree
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代码

递归

#include <iostream>
#include <cassert>

using namespace std;


/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Recursive
/// Time Complexity: O(lgn), where n is the node's number of the tree
/// Space Complexity: O(h), where h is the height of the tree
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

        assert(!p && !q); // p、q存在

        if(!root) return root;

        if(p->val < root->val && q->val < root->val) //利用平衡二叉树的性质
            return lowestCommonAncestor(root->left, p, q);
        if(p->val > root->val && q->val > root->val)
            return lowestCommonAncestor(root->right, p, q);
        return root;  //剩余三种情况：p与q在不同侧，则p、q最近公共为root；p为root，则最近公共为root;q为root，则最近公共为root
    }
};

int main() {

    return 0;
}
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思路二

#include <iostream>
#include <cassert>

using namespace std;


/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Non-Recursive
/// Time Complexity: O(lgn), where n is the node's number of the tree
/// Space Complexity: O(1)
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

        if(!root) return root;

        TreeNode* cur = root;
        while(cur) {
            if (p->val < cur->val && q->val < cur->val)
                cur = cur->left;
            else if (p->val > cur->val && q->val > cur->val)
                cur = cur->right;
            else
                return cur;
        }

        return NULL;
    }
};

int main() {

    return 0;
}
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