如何实现DFS和BFS？

• DFS（深度优先搜索）：遍历当前节点的全部子节点。再遍历同一级的节点
• BFS（广度优先搜索）：遍历兄弟节点，再遍历节点下的子节点

测试dom：

    <div id='root'>
        <span>123
            <a href="#">
                sdsd
            </a>
            <div>sdsd<a>这是一个a标签</a></div>
        </span>
        <span>456
            <p>这是一个p标签</p>
        </span>
    </div>

DFS递归实现：

    function deepTraversal(node, nodeList) {
      if (node) {
        nodeList.push(node)
        var children = node.children
        for (var i = 0; i < children.length; i++) {
          deepTraversal(children[i], nodeList)
        }
      }
      return nodeList
    }
    var root = document.getElementById('root')
    console.log(deepTraversal(root, nodeList = []))

DFS非递归实现：

   // DFS非递归实现
    function deepTraversal(node) {
      var nodeList = []
      if (node) {
        var stack = []
        stack.push(node)
        while (stack.length != 0) {
          var childrenItem = stack.pop()
          nodeList.push(childrenItem)
          var childrenList = childrenItem.children
          for (var i = childrenList.length - 1; i >= 0; i--) {
            stack.push(childrenList[i])
          }
        }
      }
      return nodeList
    }
    var root = document.getElementById('root')
    console.log(deepTraversal(root, nodeList = []))

 

BFS实现：

    function wideTraversal(node) {
      var nodes = []
      if (node != null) {
        var queue = []
        queue.unshift(node)
        while (queue.length != 0) {
          var item = queue.shift()
          nodes.push(item)
          var children = item.children
          for (var i = 0; i < children.length; i++) {
            queue.push(children[i])
          }
        }
      }
      return nodes
    }
    var root = document.getElementById('root')
    console.log(wideTraversal(root))

 

 

参考：https://www.jianshu.com/p/b4d8085e84bd