diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md
index cfd22288..cfc50c7b 100644
--- a/problems/0106.从中序与后序遍历序列构造二叉树.md
+++ b/problems/0106.从中序与后序遍历序列构造二叉树.md
@@ -580,8 +580,10 @@ tree2 的前序遍历是[1 2 3]， 后序遍历是[3 2 1]。
 
 ## 其他语言版本
 
-
 Java：
+
+106.从中序与后序遍历序列构造二叉树
+
 ```java
 class Solution {
     public TreeNode buildTree(int[] inorder, int[] postorder) {
@@ -617,8 +619,43 @@ class Solution {
 }
 ```
 
+105.从前序与中序遍历序列构造二叉树
+
+```java
+class Solution {
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+        return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
+    }
+
+    public TreeNode helper(int[] preorder, int preLeft, int preRight,
+                           int[] inorder, int inLeft, int inRight) {
+        // 递归终止条件
+        if (inLeft > inRight || preLeft > preRight) return null;
+
+        // val 为前序遍历第一个的值，也即是根节点的值
+        // idx 为根据根节点的值来找中序遍历的下标
+        int idx = inLeft, val = preorder[preLeft];
+        TreeNode root = new TreeNode(val);
+        for (int i = inLeft; i <= inRight; i++) {
+            if (inorder[i] == val) {
+                idx = i;
+                break;
+            }
+        }
+
+        // 根据 idx 来递归找左右子树
+        root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft),
+                         inorder, inLeft, idx - 1);
+        root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight,
+                         inorder, idx + 1, inRight);
+        return root;
+    }
+}
+```
+
 Python：
 105.从前序与中序遍历序列构造二叉树
+
 ```python
 # Definition for a binary tree node.
 # class TreeNode:
@@ -637,6 +674,7 @@ class Solution:
             return root
 ```
 106.从中序与后序遍历序列构造二叉树
+
 ```python
 # Definition for a binary tree node.
 # class TreeNode: