Fix：修复跨层级复制节点时，节点的富文本样式没有更新的问题

simple-mind-map/src/core/render/Render.js

@@ -1482,7 +1482,18 @@ class Render {
       node.nodeData.children.push(
         ...data.map(item => {
           const newData = simpleDeepClone(item)
-          createUidForAppointNodes([newData], true)
+          createUidForAppointNodes([newData], true, node => {
+            // 可能跨层级复制，那么富文本样式需要更新
+            if (this.mindMap.richText) {
+              // 如果设置了自定义样式那么不需要更新
+              if (
+                this.mindMap.richText.checkNodeHasCustomRichTextStyle(node.data)
+              ) {
+                return
+              }
+              node.data.resetRichText = true
+            }
+          })
           return newData
         })
       )

simple-mind-map/src/plugins/RichText.js

@@ -13,6 +13,7 @@ import {
   nodeRichTextToTextWithWrap
 } from '../utils'
 import { CONSTANTS } from '../constants/constant'
+import Node from '../core/render/node/Node'
 
 let extended = false
 
@@ -676,7 +677,7 @@ class RichText {
       'textDecoration',
       'color'
     ]
-    const nodeData = node.getData()
+    const nodeData = node instanceof Node ? node.getData() : node
     for (let i = 0; i < list.length; i++) {
       if (nodeData[list[i]] !== undefined) {
         return true

simple-mind-map/src/utils/index.js

@@ -948,7 +948,7 @@ export const addDataToAppointNodes = (appointNodes, data = {}) => {
 
 // 给指定的节点列表树数据添加uid，会修改原数据
 // createNewId默认为false，即如果节点不存在uid的话，会创建新的uid。如果传true，那么无论节点数据原来是否存在uid，都会创建新的uid
-export const createUidForAppointNodes = (appointNodes, createNewId = false) => {
+export const createUidForAppointNodes = (appointNodes, createNewId = false, handle = null) => {
   const walk = list => {
     list.forEach(node => {
       if (!node.data) {
@@ -957,6 +957,7 @@ export const createUidForAppointNodes = (appointNodes, createNewId = false) => {
       if (createNewId || isUndef(node.data.uid)) {
         node.data.uid = createUid()
       }
+      handle && handle(node)
       if (node.children && node.children.length > 0) {
         walk(node.children)
       }