在BFS

Question

我很困惑如何在穿越BFS时保持树的顶部。我接受这样的输入：

5 // total number of nodes in tree (root node will always be 1)
  // this number also tells the number of lines that'll be coming up
2 3 // left and right child of root, tree be like : 1
                                                   / \
                                                  2   3
4 5 // left and right child of 2 , tree             1
                                                   / \
                                                  2   3
                                                 / \
                                                4   5
-1 -1 // left and right child of 3 , that's null
-1 -1 // left and right child of 4
-1 -1 // left and right child of 5

这继续以上述方式，用户在BFS中输入左右孩子。但我无法理解如何实现这一目标。

我的看法是：

LinkedList<Node> list = new LinkedList<Node>
list.add(root); //initially push root
while(list.peek()){  //while there is node left in linked list
    curr = list.poll();
    curr.left = new Node(x) // x = whatever is the user input is (using nextInt)
                            // and if it's -1, i'll make left child null
    curr.right = new Node(x) // same goes for this one
    ll.add(curr);

}

最后，我需要对根节点的引用，我不知道如何获取它？而且，有没有更好的方法来实现这个任务

Answer 1

我希望下面的代码可以帮到你。

“readTree”函数用于读取树。

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         <!-- i need to swap current image with loading.gif and then display selected img from below --> 
         <!-- on click is changing currentImg (default image) var to ng-src -->
   <img ng-click="currentImg=item.imgSmall1" ng-src="{{item.imgSmall1}}">
   <img ng-click="currentImg=item.imgSmall2" ng-src="{{item.imgSmall2}}">
        ...