PattySung

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

<解題>方法一

因為是Binary Search Tree，所以可以用val的方式做思考（左小右大），如果： a. if(p.val < root.val && q.val < root.val) ：從root.left找 b. if(p.val > root.val && q.val > root.val) ：從root.right找 c. p,q分別在左、右，直接返回root


public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(p.val < root.val && q.val < root.val) {
            return lowestCommonAncestor(root.left , p ,q);
        }

        if(p.val > root.val && q.val > root.val) {
            return lowestCommonAncestor(root.right , p ,q);
        }

        return root;
    }
}

Time: O(N) Space: O(H)

<解題>方法二

和236二元樹一樣的思考方式：


public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q)  return root;
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if(left != null && right != null)   return root;
        return left != null ? left : right;
    }
}

Time: O(H) -> log(n) Space: O(H)

此題可和Leetcode-236. Lowest Common Ancestor of a Binary Tree做比較

Leetcode-235. Lowest Common Ancestor of a Binary Search Tree

<解題>方法一

<解題>方法二