Update 0101.对称二叉树.md

添加 0101.对称二叉树 Java版本，3种解法

problems/0101.对称二叉树.md

@@ -254,7 +254,105 @@ public:
 ## 其他语言版本
 
 Java：
+```Java
+    /**
+     * 递归法
+     */
+    public boolean isSymmetric1(TreeNode root) {
+        return compare(root.left, root.right);
+    }
 
+    private boolean compare(TreeNode left, TreeNode right) {
+        if (left == null && right != null) {
+            return false;
+        }
+        if (left != null && right == null) {
+            return false;
+        }
+        if (left == null && right == null) {
+            return true;
+        }
+        if (left.val != right.val) {
+            return false;
+        }
+        // 比较外侧
+        boolean compareOutside = compare(left.left, right.right);
+        // 比较内侧
+        boolean compareInside = compare(left.right, right.left);
+        return compareOutside && compareInside;
+    }
+
+    /**
+     * 迭代法
+     * 使用双端队列，相当于两个栈
+     */
+    public boolean isSymmetric2(TreeNode root) {
+        Deque<TreeNode> deque = new LinkedList<>();
+        deque.offerFirst(root.left);
+        deque.offerLast(root.right);
+        while (!deque.isEmpty()) {
+            TreeNode leftNode = deque.pollFirst();
+            TreeNode rightNode = deque.pollLast();
+            if (leftNode == null && rightNode == null) {
+                continue;
+            }
+//            if (leftNode == null && rightNode != null) {
+//                return false;
+//            }
+//            if (leftNode != null && rightNode == null) {
+//                return false;
+//            }
+//            if (leftNode.val != rightNode.val) {
+//                return false;
+//            }
+            // 以上三个判断条件合并
+            if (leftNode == null || rightNode == null || leftNode.val != rightNode.val) {
+                return false;
+            }
+            deque.offerFirst(leftNode.left);
+            deque.offerFirst(leftNode.right);
+            deque.offerLast(rightNode.right);
+            deque.offerLast(rightNode.left);
+        }
+        return true;
+    }
+
+    /**
+     * 迭代法
+     * 使用普通队列
+     */
+    public boolean isSymmetric3(TreeNode root) {
+        Queue<TreeNode> deque = new LinkedList<>();
+        deque.offer(root.left);
+        deque.offer(root.right);
+        while (!deque.isEmpty()) {
+            TreeNode leftNode = deque.poll();
+            TreeNode rightNode = deque.poll();
+            if (leftNode == null && rightNode == null) {
+                continue;
+            }
+//            if (leftNode == null && rightNode != null) {
+//                return false;
+//            }
+//            if (leftNode != null && rightNode == null) {
+//                return false;
+//            }
+//            if (leftNode.val != rightNode.val) {
+//                return false;
+//            }
+            // 以上三个判断条件合并
+            if (leftNode == null || rightNode == null || leftNode.val != rightNode.val) {
+                return false;
+            }
+            // 这里顺序与使用Deque不同
+            deque.offer(leftNode.left);
+            deque.offer(rightNode.right);
+            deque.offer(leftNode.right);
+            deque.offer(rightNode.left);
+        }
+        return true;
+    }
+```
 
 Python：