diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 3d23bea2..0c61c483 100644
--- a/notes/剑指 offer 题解.md
+++ b/notes/剑指 offer 题解.md
@@ -185,8 +185,6 @@ public boolean Find(int target, int[][] matrix) {
复杂度:O(N) + O(1)
-
-
```java
public String replaceSpace(StringBuffer str) {
int oldLen = str.length();
@@ -303,7 +301,7 @@ preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
```
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+
## 解题思路
@@ -341,11 +339,11 @@ private TreeNode reConstructBinaryTree(int[] pre, int preL, int preR, int[] in,
① 如果一个节点有右子树不为空,那么该节点的下一个节点是右子树的最左节点;
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② 否则,向上找第一个左链接指向的树包含该节点的祖先节点。
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+
```java
public class TreeLinkNode {
@@ -413,7 +411,7 @@ public int pop() {
如果使用递归求解,那么会重复计算一些子问题。例如,求 f(10) 需要计算 f(9) 和 f(8),计算 f(9) 需要计算 f(8) 和 f(7),可以看到 f(8) 被重复计算了。
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+
递归方法是将一个问题划分成多个子问题求解,动态规划也是如此,但是动态规划会把子问题的解缓存起来,避免重复求解子问题。
@@ -1086,7 +1084,7 @@ public ListNode Merge(ListNode list1, ListNode list2) {
## 题目描述
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## 解题思路
@@ -1109,7 +1107,7 @@ private boolean isSubtree(TreeNode root1, TreeNode root2) {
## 题目描述
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## 解题思路
@@ -1132,7 +1130,7 @@ private void swap(TreeNode root) {
## 题目描述
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+
## 解题思路
@@ -1242,7 +1240,7 @@ public boolean IsPopOrder(int[] pushA, int[] popA) {
例如,以下二叉树层次遍历的结果为:1,2,3,4,5,6,7
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+
## 解题思路
@@ -1338,7 +1336,7 @@ public ArrayList> Print(TreeNode pRoot) {
例如,下图是后序遍历序列 3,1,2 所对应的二叉搜索树。
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## 解题思路
@@ -1368,7 +1366,7 @@ private boolean verify(int[] sequence, int first, int last) {
下图的二叉树有两条和为 22 的路径:10, 5, 7 和 10, 12
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+
## 解题思路
@@ -1456,7 +1454,7 @@ public RandomListNode Clone(RandomListNode pHead) {
输入一棵二叉搜索树,将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点,只能调整树中结点指针的指向。
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## 解题思路
@@ -2142,7 +2140,7 @@ private void inOrder(TreeNode root, int k) {
从根结点到叶结点依次经过的结点(含根、叶结点)形成树的一条路径,最长路径的长度为树的深度。
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## 解题思路
@@ -2159,7 +2157,7 @@ public int TreeDepth(TreeNode root) {
平衡二叉树左右子树高度差不超过 1。
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## 解题思路
@@ -2559,7 +2557,7 @@ public int StrToInt(String str) {
### 二叉查找树
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+
```java
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
@@ -2571,7 +2569,7 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
### 普通二叉树
-
+
```java
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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