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	update 优化 部门树多基点展示问题 支持相同名称节点并排展示
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		@@ -43,6 +43,23 @@ public class TreeBuildUtils extends TreeUtil {
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        return TreeUtil.build(list, k, DEFAULT_CONFIG, nodeParser);
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    }
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    /**
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     * 构建树形结构
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     *
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     * @param <T>        输入节点的类型
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     * @param <K>        节点ID的类型
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     * @param parentId   顶级节点
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     * @param list       节点列表,其中包含了要构建树形结构的所有节点
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     * @param nodeParser 解析器,用于将输入节点转换为树节点
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     * @return 构建好的树形结构列表
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     */
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    public static <T, K> List<Tree<K>> build(List<T> list, K parentId, NodeParser<T, K> nodeParser) {
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        if (CollUtil.isEmpty(list)) {
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            return CollUtil.newArrayList();
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        }
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        return TreeUtil.build(list, parentId, DEFAULT_CONFIG, nodeParser);
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    }
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    /**
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     * 获取节点列表中所有节点的叶子节点
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     *
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@@ -102,11 +102,22 @@ public class SysDeptServiceImpl implements ISysDeptService, DeptService {
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        if (CollUtil.isEmpty(depts)) {
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            return CollUtil.newArrayList();
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        }
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        return TreeBuildUtils.build(depts, (dept, tree) ->
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        // 获取当前列表中每一个节点的parentId,然后在列表中查找是否有id与其parentId对应,若无对应,则表明此时节点列表中,该节点在当前列表中属于顶级节点
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        List<Tree<Long>> treeList = CollUtil.newArrayList();
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        for (SysDeptVo d : depts) {
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            Long parentId = d.getParentId();
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            SysDeptVo sysDeptVo = depts.stream().filter(it -> it.getDeptId().longValue() == parentId).findFirst().orElse(null);
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            if (sysDeptVo == null) {
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                List<Tree<Long>> trees = TreeBuildUtils.build(depts, parentId, (dept, tree) ->
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                    tree.setId(dept.getDeptId())
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                        .setParentId(dept.getParentId())
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                        .setName(dept.getDeptName())
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                        .setWeight(dept.getOrderNum()));
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                Tree<Long> tree = trees.stream().filter(it -> it.getId().longValue() == d.getDeptId()).findFirst().get();
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                treeList.add(tree);
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            }
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        }
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        return treeList;
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    }
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    /**
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