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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).”

 

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.

Companies:
LinkedIn, Facebook, Microsoft, Google, Uber

Related Topics:
Tree, Depth-First Search, Binary Search Tree, Binary Tree

Similar Questions:

Solution 1. DFS

// OJ: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
// Author: github.com/lzl124631x
// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (p->val > q->val) swap(p, q);
        function<TreeNode*(TreeNode*)> dfs = [&](TreeNode *node) -> TreeNode* {
            if (!node) return nullptr;
            if (node->val >= p->val && node->val <= q->val) return node;
            if (node->val < p->val) return dfs(node->right);
            return dfs(node->left);
        };
        return dfs(root);
    }
};

Solution 2. DFS

// OJ: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
// Author: github.com/lzl124631x
// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root->val > p->val && root->val > q->val) return lowestCommonAncestor(root->left, p, q);
        if (root->val < p->val && root->val < q->val) return lowestCommonAncestor(root->right, p, q);
        return root;
    }
};