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235. Lowest Common Ancestor of a Binary Search Tree

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:

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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:

img

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.

Solution:

  1. 当前节点是空节点 不用判断 p and q will exist in the BST.

  2. p和q都在左子树: 返回递归左子树的结果

  3. p和q都在右子树: 返回递归右子树的结果

  4. 其他:

4.1 p和q分别在左右子树

4.2 当前节点是p

4.3 当前节点是q

返回当前节点

搜索树的好处: 提前剪枝

/**
 * 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) {
        int cur = root.val;
        if (p.val < cur && q.val < cur){
            return lowestCommonAncestor(root.left, p, q);
        }


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

        return root;

    }
}

// TC: O(n)
// SC: O(n)

/**
 * 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) {
        // base case
        if (root == null){
            return null;
        }

        if (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;
        }

        if (left == null && right == null){
            return null;
        }

        if (left != null){
            return left;
        }else{
            return right;
        }
    }
}
/*
    case 1 : root.left == null && root.right == null -> return null
    case 2 : root.left != null || root.right != null -> return not null side like root.left or root.right
    case 3 : root.left && root.right != null return root;
*/
// TC: O(n)
// SC: O(n)

Binary Search Tree (BST): Due to the inherent property of a BST (left child is smaller, and right child is larger), finding the LCA is relatively easier and can be done in a more straightforward manner.

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