diff --git a/java-leetcode/README.md b/java-leetcode/README.md
index b3d88bb0d626b6c980f980c798dc3bd27f856f74..2af2e3cad1f7aaf98fd2b97f7366e39e5e2fbe98 100644
--- a/java-leetcode/README.md
+++ b/java-leetcode/README.md
@@ -10,3 +10,4 @@
 * [输出二叉树中所有从根到叶子的路径](./src/main/java/io/github/dreamylost/Leetcode_257_DFS.java)
 * [BiNode](./src/main/java/io/github/dreamylost/Leetcode_Interview_17_12.java)
 * [二叉搜索树的最近公共祖先](./src/main/java/io/github/dreamylost/Leetcode_JZ_68_1.java)
+* [二叉树的最近公共祖先](./src/main/java/io/github/dreamylost/Leetcode_JZ_68_2.java)
diff --git a/java-leetcode/src/main/java/io/github/dreamylost/Leetcode_JZ_68_2.java b/java-leetcode/src/main/java/io/github/dreamylost/Leetcode_JZ_68_2.java
new file mode 100644
index 0000000000000000000000000000000000000000..d4f061f84c9554978d17dee0ed27486a13a71034
--- /dev/null
+++ b/java-leetcode/src/main/java/io/github/dreamylost/Leetcode_JZ_68_2.java
@@ -0,0 +1,115 @@
+/* Licensed under Apache-2.0 @梦境迷离 */
+package io.github.dreamylost;
+
+import java.util.*;
+
+/**
+ * 剑指 Offer 68 - II. 二叉树的最近公共祖先
+ *
+ * @author 梦境迷离 dreamylost
+ * @version v1.0
+ * @since 2020-06-26
+ */
+public class Leetcode_JZ_68_2 {
+
+    /**
+     * 7 ms,100.00% 41.8 MB,100.00%
+     *
+     * @param root
+     * @param p
+     * @param q
+     * @return
+     */
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        //当p或q本身就是根节点时，公共就是p或q本身，因为根之上没有父节点
+        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 null;
+        }
+        if (left != null && right != null) {
+            return root;
+        }
+        return left == null ? right : left;
+    }
+
+    /**
+     * author 2hp 暴力 7 ms,100.00% 40.9 MB,100.00%
+     *
+     * @param root
+     * @param p
+     * @param q
+     * @return
+     */
+    public TreeNode lowestCommonAncestor2(TreeNode root, TreeNode p, TreeNode q) {
+        //如果从p开始向下找有q则返回p
+        if (dfs(p, q)) return p;
+        //如果从q开始向下找有p返回q
+        if (dfs(q, p)) return q;
+        //否则
+        return help(root, p, q);
+    }
+
+    //方法：在root向下寻找是否有search节点
+    public boolean dfs(TreeNode root, TreeNode search) {
+        if (root == null) return false;
+        if (root.val == search.val) return true;
+        else return dfs(root.left, search) || dfs(root.right, search);
+    }
+
+    //方法：1.root子节点中分别含有p、q
+    //2.root单边含有p、q
+    public TreeNode help(TreeNode root, TreeNode p, TreeNode q) {
+        //如果root.left下有p并且root.right下有q返回root；相反也是一样返回root
+        if ((dfs(root.left, p) && dfs(root.right, q)) || dfs(root.left, q) && dfs(root.right, p)) {
+            return root;
+        } else if (dfs(root.left, p) && dfs(root.left, q))
+            return help(root.left, p, q); //否则p,q就是在同一边，递归调用即可
+        else return help(root.right, p, q);
+    }
+
+    /**
+     * author sdwwld
+     *
+     * <p>15 ms,7.16% 40 MB,100.00%
+     *
+     * <p>非递归 BFS
+     *
+     * @param root
+     * @param p
+     * @param q
+     * @return
+     */
+    public TreeNode lowestCommonAncestor3(TreeNode root, TreeNode p, TreeNode q) {
+        //记录遍历到的每个节点的父节点
+        Map<TreeNode, TreeNode> parentMap = new HashMap<>();
+        Queue<TreeNode> queue = new LinkedList<>();
+        parentMap.put(root, null);
+        queue.add(root);
+        //直到两个节点都找到为止
+        while (!parentMap.containsKey(p) || !parentMap.containsKey(q)) {
+            TreeNode node = queue.poll();
+            if (node != null && node.left != null) {
+                parentMap.put(node.left, node);
+                queue.add(node.left);
+            }
+            if (node != null && node.right != null) {
+                parentMap.put(node.right, node);
+                queue.add(node.right);
+            }
+        }
+
+        Set<TreeNode> ancestorsSet = new HashSet<>();
+        //记录下p和他的祖先节点，从p节点开始一直到根节点
+        while (p != null) {
+            ancestorsSet.add(p);
+            p = parentMap.get(p);
+        }
+        //查看p和他的祖先节点是否包含q节点，如果不包含再看是否包含q的父节点
+        while (!ancestorsSet.contains(q)) q = parentMap.get(q);
+        return q;
+    }
+}