diff --git a/notes/68. 树中两个节点的最低公共祖先.md b/notes/68. 树中两个节点的最低公共祖先.md
index 9113ba6b..e75dcea4 100644
--- a/notes/68. 树中两个节点的最低公共祖先.md	
+++ b/notes/68. 树中两个节点的最低公共祖先.md	
@@ -9,9 +9,9 @@
 
 ### 解题思路
 
-在二叉查找树中，两个节点 p, q 的公共祖先 root 满足 root.val \>= p.val && root.val \<= q.val。
+在二叉查找树中，设 p.val \<= q.val，从根节点往下查找，第一个满足 root.val \>= p.val && root.val \<= q.val 的节点 root 就是 p、q 的最低公共祖先。
 
-<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/047faac4-a368-4565-8331-2b66253080d3.jpg" width="250"/> </div><br>
+<div align="center"> <img src="pics/offer-bst-lca.png" width="460px"/> </div><br>
 
 ```java
 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
@@ -35,7 +35,7 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 
 在左右子树中查找是否存在 p 或者 q，如果 p 和 q 分别在两个子树中，那么就说明根节点就是最低公共祖先。
 
-<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/d27c99f0-7881-4f2d-9675-c75cbdee3acd.jpg" width="250"/> </div><br>
+<div align="center"> <img src="pics/offer-tree-lca.png" width="460px"/> </div><br>
 
 ```java
 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
diff --git a/notes/pics/offer-bst-lca.png b/notes/pics/offer-bst-lca.png
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diff --git a/notes/pics/offer-tree-lca.png b/notes/pics/offer-tree-lca.png
new file mode 100644
index 00000000..2c483d41
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