Question

Formatted question description: https://leetcode.ca/all/235.html

 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).”

 Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

       _______6______
      /              \
  ___2__          ___8__
 /      \        /      \
0      _4       7       9
      /  \
     3   5

 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.

 Note:

 All of the nodes' values will be unique.
 p and q are different and both values will exist in the BST.

 @tag-tree

Algorithm

If the value of the root node is greater than the larger value between p and q, indicating that both p and q are in the left subtree, then we enter the left child node of the root node to continue recursion,

If the root node is less than the smaller value between p and q, indicating that both p and q are in the right subtree, then we enter the right child node of the root node to continue recursion,

If neither, it means that the current root node is the smallest common parent node and return directly.

Code

Java

import java.util.ArrayList;
import java.util.List;

public class Lowest_Common_Ancestor_of_a_Binary_Search_Tree {

    public static void main(String[] args) {
        Lowest_Common_Ancestor_of_a_Binary_Search_Tree out = new Lowest_Common_Ancestor_of_a_Binary_Search_Tree();
        Solution s = out.new Solution();

        TreeNode root = new TreeNode(6);
        root.left = new TreeNode(2);
        root.left.left = new TreeNode(0);
        root.left.right = new TreeNode(4);

        System.out.println(s.lowestCommonAncestor(root, root.left, root.left.right).val);
    }

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */

    class Solution_noExtraSpace {
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
            if (root == null) {
                return root;
            }
            if (root.val > Math.max(p.val, q.val)) {
                return lowestCommonAncestor(root.left, p, q);
            } else if (root.val < Math.min(p.val, q.val)) {
                return lowestCommonAncestor(root.right, p, q);
            } else {
                return root;
            }
        }
    }

    class Solution_iteration {
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
            while (true) {
                if (root.val > Math.max(p.val, q.val)) root = root.left;
                else if (root.val < Math.min(p.val, q.val)) root = root.right;
                else break;
            }
            return root;
        }
    }

    class Solution {
        List<TreeNode> pList;
        List<TreeNode> qList;

        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

            backTrack(root, p, q, new ArrayList<TreeNode>());

            TreeNode result = null;
            for (int i = 0; i < Math.min(pList.size(), qList.size()); i++) {
//                if (pList.get(i).val == qList.get(i).val) {
                if (pList.get(i) == qList.get(i)) { // compare hash code
                    result = pList.get(i);
                }
            }

            return result;
        }

        private void backTrack(TreeNode root, TreeNode p, TreeNode q, ArrayList<TreeNode> list) {
            if (root == null) {
                return;
            }

            ArrayList<TreeNode> newList = new ArrayList<>(list);
            newList.add(root);

            // @note:@memorize: final list should contain current root value, in case p is child of q
            if (root.val == p.val) {
                // found p
                pList = newList;
                // @note:@memorize: cannot stop here, q could be child of p
//                return; // no need to continue recursion
            }

            if (root.val == q.val) {
                qList = newList;
//                return;
            }

            backTrack(root.left, p, q, newList);
            backTrack(root.right, p, q, newList);
        }
    }
}

All Problems

All Solutions