如何判断是平衡二叉树

Posted 2020-10-27 提拉没有米苏

依据：平衡二叉树的左右子树高度差不超过1

　　private boolean isBalanced = true;
    
    public boolean isbalanced_solutions(TreeNode root) {
        height(root);
        return isBalanced;
    }

    private int height(TreeNode root) {
        if(null == root) {
            return 0;
        }
        int left = height(root.left);
        int right = height(root.right);
        if(Math.abs(left - right) > 1) {
            isBalanced = false;
        }
        return 1+ Math.max(left, right);
    }