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[Leetcode]235. Lowest Common Ancestor of a Binary Search Tree(C++)

题目描述

题目链接:235. Lowest Common Ancestor of a Binary Search Tree

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).”

例子

例子 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.

例子 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.

例子 3

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

Constraints

  • The number of nodes in the tree is in the range [2, 10^5].
  • -10^9 <= Node.val <= 10^9
  • All Node.val are unique.
  • p != q
  • p and q will exist in the BST.

解题思路

由于题目说了是二叉搜索树,所以通过根结点值和两个节点值很容易可以判断两个节点是在根结点的两侧还是同一侧。如果是两侧则返回根结点,否则递归搜索该侧即可,代码如下:

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/**
* 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 == p || root == q) return root;
if (p->val < q->val) {
if (p->val < root->val && root->val < q->val)
return root;
else if (q->val < root->val) {
return lowestCommonAncestor(root->left, p, q);
} else {
return lowestCommonAncestor(root->right, p, q);
}
} else {
if (p->val > root->val && root->val > q->val) {
return root;
} else if (p->val < root->val) {
return lowestCommonAncestor(root->left, p, q);
} else {
return lowestCommonAncestor(root->right, p, q);
}
}
}
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(h)

GitHub 代码同步地址: 235.LowestCommonAncestorOfABinarySearchTree.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions