ps：练习来自力扣

给定一个二叉树，判断它是否是平衡二叉树

// 定义二叉树节点类
class TreeNode {int val;TreeNode left;TreeNode right;TreeNode() {}TreeNode(int val) { this.val = val; }TreeNode(int val, TreeNode left, TreeNode right) {this.val = val;this.left = left;this.right = right;}
}class Solution {public boolean isBalanced(TreeNode root) {return getHeight(root) != -1;}private int getHeight(TreeNode node) {// 如果当前节点为空，高度为 0if (node == null) {return 0;}// 递归计算左子树的高度int leftHeight = getHeight(node.left);// 如果左子树已经不平衡，直接返回 -1if (leftHeight == -1) {return -1;}// 递归计算右子树的高度int rightHeight = getHeight(node.right);// 如果右子树已经不平衡，直接返回 -1if (rightHeight == -1) {return -1;}// 判断当前节点的左右子树高度差是否超过 1if (Math.abs(leftHeight - rightHeight) > 1) {return -1;}// 如果当前节点平衡，返回其高度（左右子树最大高度加 1）return Math.max(leftHeight, rightHeight) + 1;}
}