Update 0110.平衡二叉树.md

添加 0110.平衡二叉树 Java版本，3种解法

problems/0110.平衡二叉树.md

@@ -354,7 +354,146 @@ public:
 ## 其他语言版本
 
 Java：
+```Java
+class Solution {
+   /**
+     * 递归法
+     */
+    public boolean isBalanced(TreeNode root) {
+        return getHeight(root) != -1;
+    }
 
+    private int getHeight(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        int leftHeight = getHeight(root.left);
+        if (leftHeight == -1) {
+            return -1;
+        }
+        int rightHeight = getHeight(root.right);
+        if (rightHeight == -1) {
+            return -1;
+        }
+        // 左右子树高度差大于1，return -1表示已经不是平衡树了
+        if (Math.abs(leftHeight - rightHeight) > 1) {
+            return -1;
+        }
+        return Math.max(leftHeight, rightHeight) + 1;
+    }
+}
+
+class Solution {
+   /**
+     * 迭代法，效率较低，计算高度时会重复遍历
+     * 时间复杂度：O(n^2)
+     */
+    public boolean isBalanced(TreeNode root) {
+        if (root == null) {
+            return true;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        TreeNode pre = null;
+        while (root!= null || !stack.isEmpty()) {
+            while (root != null) {
+                stack.push(root);
+                root = root.left;
+            }
+            TreeNode inNode = stack.peek();
+            // 右结点为null或已经遍历过
+            if (inNode.right == null || inNode.right == pre) {
+                // 比较左右子树的高度差，输出
+                if (Math.abs(getHeight(inNode.left) - getHeight(inNode.right)) > 1) {
+                    return false;
+                }
+                stack.pop();
+                pre = inNode;
+                root = null;// 当前结点下，没有要遍历的结点了
+            } else {
+                root = inNode.right;// 右结点还没遍历，遍历右结点
+            }
+        }
+        return true;
+    }
+
+    /**
+     * 层序遍历，求结点的高度
+     */
+    public int getHeight(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        Deque<TreeNode> deque = new LinkedList<>();
+        deque.offer(root);
+        int depth = 0;
+        while (!deque.isEmpty()) {
+            int size = deque.size();
+            depth++;
+            for (int i = 0; i < size; i++) {
+                TreeNode poll = deque.poll();
+                if (poll.left != null) {
+                    deque.offer(poll.left);
+                }
+                if (poll.right != null) {
+                    deque.offer(poll.right);
+                }
+            }
+        }
+        return depth;
+    }
+}
+
+class Solution {
+   /**
+     * 优化迭代法，针对暴力迭代法的getHeight方法做优化，利用TreeNode.val来保存当前结点的高度，这样就不会有重复遍历
+     * 获取高度算法时间复杂度可以降到O(1)，总的时间复杂度降为O(n)。
+     * <p>
+     * 时间复杂度：O(n)
+     */
+    public boolean isBalanced(TreeNode root) {
+        if (root == null) {
+            return true;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        TreeNode pre = null;
+        while (root != null || !stack.isEmpty()) {
+            while (root != null) {
+                stack.push(root);
+                root = root.left;
+            }
+            TreeNode inNode = stack.peek();
+            // 右结点为null或已经遍历过
+            if (inNode.right == null || inNode.right == pre) {
+                // 输出
+                if (Math.abs(getHeight(inNode.left) - getHeight(inNode.right)) > 1) {
+                    return false;
+                }
+                stack.pop();
+                pre = inNode;
+                root = null;// 当前结点下，没有要遍历的结点了
+            } else {
+                root = inNode.right;// 右结点还没遍历，遍历右结点
+            }
+        }
+        return true;
+    }
+
+    /**
+     * 求结点的高度
+     */
+    public int getHeight(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        int leftHeight = root.left != null ? root.left.val : 0;
+        int rightHeight = root.right != null ? root.right.val : 0;
+        int height = Math.max(leftHeight, rightHeight) + 1;
+        root.val = height;// 用TreeNode.val来保存当前结点的高度
+        return height;
+    }
+}
+LeetCode题解链接：https://leetcode-cn.com/problems/balanced-binary-tree/solution/110-ping-heng-er-cha-shu-di-gui-fa-bao-l-yqr3/
+```
 
 Python：