JS 两树交集

Posted 2022-11-27 码小余の博客

求两个树形结构的交集

思路是遍历递归第一棵树，在遍历内部再次遍历递归第二棵树，添加层级，用层级和id进行比较，如果相等就添加不用删除的标记

然后在下面在遍历递归tree2，删除没有exist的节点，完成

const tree1 = [
  
    id: 1,
    name: 'aaa',
  ,
  
    id: 2,
    name: 'bbb',
    children: [
      
        id: 3,
        name: 'ccc'
      ,
      
        id: 4,
        name: 'ddd'
      
    ]
  ,
  
    id: 5,
    name: 'eee',
    children: [
      
        id: 6,
        name: 'fff'
      ,
      
        id: 7,
        name: 'ggg',
        children: [
          
            id: 8,
            name: 'hhh',
          ,
          
            id: 9,
            name: 'iii',
          
        ]
      
    ]
  ,
]

const tree2 = [
  
    id: 1,
    name: 'aaa',
  ,
  
    id: 2,
    name: 'bbb',
    children: [
      
        id: 3,
        name: 'ccc'
      ,
      
        id: 44,
        name: 'jjj'
      
    ]
  ,
  
    id: 5,
    name: 'eee',
    children: [
      
        id: 6,
        name: 'fff'
      ,
      
        id: 7,
        name: 'ggg',
        children: [
          
            id: 100,
            name: 'zzz'
          ,
          
            id: 8,
            name: 'hhh',
          ,
          
            id: 99,
            name: 'kkk',
          
        ]
      
    ]
  ,
]

/**
 * 递归遍历树节点
 * @param * nodes 树节点数据
 */
function f (nodes = [], nodes2 = [], level = 0, level2 = 0) 
  level++
  nodes.forEach((node) => 
    node.level = level
    // console.log(node);

    // 循环遍历tree2
    function f2 (nodes2 = [], level2 = 0) 
      // console.log("nodes2: ", nodes2);
      level2++
      nodes2.forEach((node2) => 
        node2.level = level2
        // console.log(node2);

        if (node.level === node2.level && node.id === node2.id) 
          node2.exist = true
        

        var mapNode2 = new Map()
        for (var k2 in node2) 
          mapNode2.set(k2, node2[k2])
        
        if (mapNode2.has("children")) f2(mapNode2.get("children"), level2)
      )
    
    f2(nodes2)

    var mapNode = new Map()
    for (var k in node) 
      mapNode.set(k, node[k])
    
    if (mapNode.has("children")) f(mapNode.get("children"), nodes2, level)
  )


f(tree1, tree2)

function filterFn (nodes = []) 
  for (let i = nodes.length - 1; i >= 0; i--) 
    var mapNode = new Map()
    for (var k in nodes[i]) 
      mapNode.set(k, nodes[i][k])
    
    if (!mapNode.has("exist")) 
      nodes.splice(i, 1)
    

    if (mapNode.has("children")) filterFn(mapNode.get("children"))
  


filterFn(tree2)
// 两树交集如下结果
console.log(tree2);

emm，不会写，就写成这样了，有大佬会写的帮忙看看有没有其他的写法~