Merge pull request #95 from Joshua-Lu/patch-11

更新 0106.从中序与后序遍历序列构造二叉树 Java版本

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -585,46 +585,38 @@ Java：
 ```java
 class Solution {
     public TreeNode buildTree(int[] inorder, int[] postorder) {
-        if (inorder.length == 0 || postorder.length == 0) return null;
+        return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
-        return buildTree(inorder, postorder,0,inorder.length,0,postorder.length);
     }
-
+    public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
-    private TreeNode buildTree(int[] inorder, int[] postorder, int infrom, int into,
+                               int[] postorder, int postLeft, int postRight) {
-                                      int postfrom, int postto) {
+        // 没有元素了
-        if (postfrom == postto) return null;
+        if (inRight - inLeft < 1) {
-
+            return null;
-        int rootValue = postorder[postto - 1];
+        }
-        TreeNode root = new TreeNode(rootValue);
+        // 只有一个元素了
-
+        if (inRight - inLeft == 1) {
-        if (postfrom + 1 == postto) return root;
+            return new TreeNode(inorder[inLeft]);
-
+        }
-        int splitNum = postto - 1;
+        // 后序数组postorder里最后一个即为根结点
-        for (int i = infrom; i < into; i++) {
+        int rootVal = postorder[postRight - 1];
-            if (inorder[i] == rootValue) {
+        TreeNode root = new TreeNode(rootVal);
-                splitNum = i;
+        int rootIndex = 0;
-                break;
+        // 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
-        int inLeftBegin = infrom;
+        root.left = buildTree1(inorder, inLeft, rootIndex,
-        int inLeftEnd = splitNum;
+                postorder, postLeft, postLeft + (rootIndex - inLeft));
-        int inRightBegin = splitNum + 1;
+        root.right = buildTree1(inorder, rootIndex + 1, inRight,
-        int inRightEnd = into;
+                postorder, postLeft + (rootIndex - inLeft), postRight - 1);
-
-        int postLeftBegin = postfrom;
-        int postLeftEnd = postLeftBegin + (splitNum - inLeftBegin);
-        int postRightBegin = postLeftBegin + (splitNum - inLeftBegin);
-        int postRightEnd = postto - 1;
-
-        root.left = buildTree(inorder,postorder,inLeftBegin,inLeftEnd,postLeftBegin,postLeftEnd);
-        root.right = buildTree(inorder,postorder,inRightBegin,inRightEnd,postRightBegin,postRightEnd);
-
         return root;
     }
 }
 ```
 
-
 Python：