From a0e346b02282ac515b8c453abce795eff2189519 Mon Sep 17 00:00:00 2001
From: Joshua <47053655+Joshua-Lu@users.noreply.github.com>
Date: Fri, 14 May 2021 00:56:39 +0800
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添加 0106.从中序与后序遍历序列构造二叉树 Java版本
---
 ...序与后序遍历序列构造二叉树.md | 34 +++++++++++++++++++
 1 file changed, 34 insertions(+)

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 f1f30b71..5533a818 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
@@ -582,7 +582,41 @@ tree2 的前序遍历是[1 2 3]， 后序遍历是[3 2 1]。
 
 
 Java：
+```java
+class Solution {
+    public TreeNode buildTree(int[] inorder, int[] postorder) {
+        return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
+    }
 
+    public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
+                               int[] postorder, int postLeft, int postRight) {
+        // 没有元素了
+        if (inRight - inLeft < 1) {
+            return null;
+        }
+        // 只有一个元素了
+        if (inRight - inLeft == 1) {
+            return new TreeNode(inorder[inLeft]);
+        }
+        // 后序数组postorder里最后一个即为根结点
+        int rootVal = postorder[postRight - 1];
+        TreeNode root = new TreeNode(rootVal);
+        int rootIndex = 0;
+        // 根据根结点的值找到该值在中序数组inorder里的位置
+        for (int i = inLeft; i < inRight; i++) {
+            if (inorder[i] == rootVal) {
+                rootIndex = i;
+            }
+        }
+        // 根据rootIndex划分左右子树
+        root.left = buildTree1(inorder, inLeft, rootIndex,
+                postorder, postLeft, postLeft + (rootIndex - inLeft));
+        root.right = buildTree1(inorder, rootIndex + 1, inRight,
+                postorder, postLeft + (rootIndex - inLeft), postRight - 1);
+        return root;
+    }
+}
+```
 
 Python：