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[8/12] 235. Lowest Common Ancestor of a Binary Search Tree

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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, 105].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • p != q
  • p and q will exist in the BST.
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학부유학생님의 댓글

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Runtime: 146 ms, faster than 25.90% of Python3 online submissions for Lowest Common Ancestor of a Binary Search Tree.
Memory Usage: 18.8 MB, less than 22.99% of Python3 online submissions for Lowest Common Ancestor of a Binary Search Tree.
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if not root: return None
        if root == p or root == q: return root
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        
        if left != None and right != None: return root
        
        if left != None: return left
        if right != None: return right

재민재민님의 댓글

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Runtime: 34 ms, faster than 86.93% of C++ online submissions for Lowest Common Ancestor of a Binary Search Tree.
Memory Usage: 23.2 MB, less than 91.99% of C++ online submissions for Lowest Common Ancestor of a Binary Search Tree.

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        int s_val = min(p->val, q->val);
        int l_val = max(p->val, q->val);
        
        if(s_val == root->val || root->val == l_val)
            return root;
        else if(s_val < root->val && root->val < l_val)
            return root;
        else if(s_val < root->val && l_val < root->val )
            return lowestCommonAncestor(root->left, p, q);
        else if(s_val > root->val && l_val > root->val )
            return lowestCommonAncestor(root->right, p, q);
        
        return NULL;
            
    }
};
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