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

Given a binary search tree (BST), find the lowest common ancestor (LCA) node 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)."

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.

Example 3:

Input: root = [2,1], p = 2, q = 1
Output: 2

Constraints:
  • The number of nodes in the tree is in the range [ 2 , 1 0 5 ] [2, 10^5] [2,105].
  • − 1 0 9 < = N o d e . v a l < = 1 0 9 -10^9 <= Node.val <= 10^9 −109<=Node.val<=109
  • All Node.val are unique.
  • p != q
  • p and q will exist in the BST.

From: LeetCode

Link: 233. Number of Digit One


Solution:

Ideas:
  • Start from the root and keep traversing the tree.
  • Depending on the values of p and q relative to the current node, move left or right.
  • If you find a node where p and q lie on different sides (or one of them is equal to the current node), that node is the LCA.
Code:
c 复制代码
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    while (root != NULL) {
        // If both p and q are greater than root, LCA lies in the right subtree
        if (p->val > root->val && q->val > root->val) {
            root = root->right;
        }
        // If both p and q are smaller than root, LCA lies in the left subtree
        else if (p->val < root->val && q->val < root->val) {
            root = root->left;
        }
        // We have found the split point, i.e. the LCA node
        else {
            return root;
        }
    }
    return NULL; // this line will never be reached if p and q are guaranteed to be in the tree
}
相关推荐
QuantumStack4 分钟前
【C++ 真题】P1104 生日
开发语言·c++·算法
写个博客41 分钟前
暑假算法日记第一天
算法
绿皮的猪猪侠43 分钟前
算法笔记上机训练实战指南刷题
笔记·算法·pta·上机·浙大
hie988941 小时前
MATLAB锂离子电池伪二维(P2D)模型实现
人工智能·算法·matlab
杰克尼2 小时前
BM5 合并k个已排序的链表
数据结构·算法·链表
.30-06Springfield2 小时前
决策树(Decision tree)算法详解(ID3、C4.5、CART)
人工智能·python·算法·决策树·机器学习
我不是哆啦A梦2 小时前
破解风电运维“百模大战”困局,机械版ChatGPT诞生?
运维·人工智能·python·算法·chatgpt
xiaolang_8616_wjl2 小时前
c++文字游戏_闯关打怪
开发语言·数据结构·c++·算法·c++20
small_wh1te_coder3 小时前
硬件嵌入式学习路线大总结(一):C语言与linux。内功心法——从入门到精通,彻底打通你的任督二脉!
linux·c语言·汇编·嵌入式硬件·算法·c
挺菜的3 小时前
【算法刷题记录(简单题)002】字符串字符匹配(java代码实现)
java·开发语言·算法