Question

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

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.

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

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)) { // no =, so root is p or q is in else block
                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);
        }
    }
}

############

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

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

// OJ: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
// 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);
    }
};

# 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':
        while 1:
            if root.val < min(p.val, q.val): # no =, so root is p or q is in else block
                root = root.right
            elif root.val > max(p.val, q.val):
                root = root.left
            else:
                return root

############

class Solution(object):
  def lowestCommonAncestor(self, root, p, q):
    """
    :type root: TreeNode
    :type p: TreeNode
    :type q: TreeNode
    :rtype: TreeNode
    """
    a, b = sorted([p.val, q.val])
    while not a <= root.val <= b:
      if a > root.val:
        root = root.right
      else:
        root = root.left
    return root

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val   int
 *     Left  *TreeNode
 *     Right *TreeNode
 * }
 */

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
	for {
		if root.Val < p.Val && root.Val < q.Val {
			root = root.Right
		} else if root.Val > p.Val && root.Val > q.Val {
			root = root.Left
		} else {
			return root
		}
	}
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function lowestCommonAncestor(
    root: TreeNode | null,
    p: TreeNode | null,
    q: TreeNode | null,
): TreeNode | null {
    while (root) {
        if (root.val > p.val && root.val > q.val) {
            root = root.left;
        } else if (root.val < p.val && root.val < q.val) {
            root = root.right;
        } else {
            return root;
        }
    }
}

235 - Lowest Common Ancestor of a Binary Search Tree

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