logseq_notes/pages/OJ notes/pages/Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree.md

Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree

2022-07-08 11:53

Algorithms:

#algorithm #binary_search #DFS

Data structures:

#DS #binary_tree #binary_search_tree

Difficulty:

#coding_problems #difficulty_easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

  ---

Problem

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).”

Examples

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.

Thoughts

[!summary] This is a #binary_search

Because of the features of BST, the maximum val of a left sub-tree is smaller than node, so the valid LCA must meet this:

root->val >= small && root->val <= big

otherwise, search the left or right subtree.

Solution

/**
 * 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) {
  // DFS-like recursion
  // Base cases
  int big = max(q->val, p->val);
  int small = min(q->val, p->val);

  if (root->val >= small && root->val <= big) {
    return root;
  } else if (root->val > big) {
    return lowestCommonAncestor(root->left, p, q);
  } else {
    return lowestCommonAncestor(root->right, p, q);
  }
}
};