二刷235

Author: diguage

docs/0235-lowest-common-ancestor-of-a-binary-search-tree.adoc

@@ -58,10 +58,26 @@ image::images/0235-06.png[{image_attr}]
 
 
 [[src-0235]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0235_LowestCommonAncestorOfABinarySearchTree.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0235_LowestCommonAncestorOfABinarySearchTree_2.java[tag=answer]
+----
+--
+====
 
 作为这道题的延伸，继续看 xref:0236-lowest-common-ancestor-of-a-binary-tree.adoc[236. Lowest Common Ancestor of a Binary Tree]。
 

logbook/202406.adoc

@@ -697,6 +697,10 @@
 |{doc_base_url}/0724-find-pivot-index.adoc[题解]
 |✅ 前缀和
 
+|{counter:codes}
+|{leetcode_base_url}/lowest-common-ancestor-of-a-binary-search-tree/[235. Lowest Common Ancestor of a Binary Search Tree^]
+|{doc_base_url}/0235-lowest-common-ancestor-of-a-binary-search-tree.adoc[题解]
+|✅ 二叉搜索树左大右小，祖先在中间的特性，可以快速去除不必要的判断。
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_0235_LowestCommonAncestorOfABinarySearchTree.java

@@ -10,6 +10,9 @@ public class _0235_LowestCommonAncestorOfABinarySearchTree {
   // tag::answer[]
   /**
    * 要充分利用二叉搜索树的特性
+   *
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-06-24 18:21:20
    */
   public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
     TreeNode result = root;

src/main/java/com/diguage/algo/leetcode/_0235_LowestCommonAncestorOfABinarySearchTree_2.java

@@ -0,0 +1,22 @@
+package com.diguage.algo.leetcode;
+
+import com.diguage.algo.util.TreeNode;
+
+public class _0235_LowestCommonAncestorOfABinarySearchTree_2 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-18 22:02:24
+   */
+  public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+    if ((p.val <= root.val && root.val <= q.val)
+      || (q.val <= root.val && root.val <= p.val)) {
+      return root;
+    } else if (p.val <= root.val && q.val <= root.val) {
+      return lowestCommonAncestor(root.left, p, q);
+    } else {
+      return lowestCommonAncestor(root.right, p, q);
+    }
+  }
+  // end::answer[]
+}