请教用 Java 如何写一个用来将 leetcode 上给出的[4,1,null,2,null,3]这种形式的树,转换为一颗真的树的根节点的函数? - V2EX
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Newyorkcity
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请教用 Java 如何写一个用来将 leetcode 给出的[4,1,null,2,null,3]这种形式的树,转换为一颗真的树的根节点的函数?

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  •   Newyorkcity 2020-04-14 18:18:05 +08:00 1541 次点击
    这是一个创建于 2019 天前的主题,其中的信息可能已经有所发展或是发生改变。
    看似是层次遍历,但实际不是,不是之处在于,比如这颗树深度有 4 层,但第二层(共 2 个节点)中右边那个节点为 null,那显然这个节点的第三层的两个子节点也为 null,但层次遍历中这两个 null 是会出现的,而这种形式里不会出现。

    我的想法是用队列往里放 node.left node.right 跟着这个形式,但遇到的问题是 Java 是值传递且没有引用传递的方式(吗?),因而往队列里放 node.left 再取出后,实际上没能改到 node.left 。。

    谢谢
    第 1 条附言    2020-04-14 19:44:43 +08:00

    给出一个我的实现,目前用了几个都对~

     public static TreeNode build(Integer[] vals) { int valid_vals_size = vals.length; Queue<TreeNode> queue = new LinkedList<>(); LinkedList<Integer> level_order_traveresal = new LinkedList<>(Arrays.asList(vals)); TreeNode root = new TreeNode(vals[0]); queue.add(root); valid_vals_size--; int index = 0; while (true) { TreeNode poll = queue.poll(); int left_index = 2 * index + 1; int right_index = 2 * index + 2; if (poll == null) { level_order_traveresal.add(left_index, null); level_order_traveresal.add(right_index, null); } else { if (left_index >= level_order_traveresal.size()) { break; } Integer left_node = level_order_traveresal.get(left_index); if (left_node != null) { poll.left = new TreeNode(left_node); valid_vals_size--; } queue.add(poll.left); if (right_index >= level_order_traveresal.size()) break; Integer right_node = level_order_traveresal.get(right_index); if (right_node != null) { poll.right = new TreeNode(right_node); valid_vals_size--; } queue.add(poll.right); } index++; if (valid_vals_size <= 0) break; } return root; } 
    7 条回复    2020-04-14 20:41:35 +08:00
    kkkkkrua
        1
    kkkkkrua  
       2020-04-14 18:43:29 +08:00
    Java 是值传递且没有引用传递
    反了吧?
    kkkkkrua
        2
    kkkkkrua  
       2020-04-14 18:48:34 +08:00
    @kkkkkrua #1 错了,看成了 C#的值类型,引用类型。。
    forestn
        3
    forestn  
       2020-04-14 19:03:34 +08:00 via iPhone
    深度优先递归吧?
    Newyorkcity
        4
    Newyorkcity  
    OP
       2020-04-14 19:15:11 +08:00
    @forestn 具体一些?
    zmxnv123
        5
    zmxnv123  
       2020-04-14 19:31:34 +08:00
    用一个队列存放当前层的 Node.
    luckyrayyy
        6
    luckyrayyy  
       2020-04-14 19:36:56 +08:00
    没看懂你的意思,都为 null 了还便利他的子节点干啥?
    xxdd
        7
    xxdd  
       2020-04-14 20:41:35 +08:00
    /**
    * 10 0
    * / \
    * 5 -3 1-2
    * / \ \
    * 3 2 11 3-6
    * / \ \
    * 3 -2 1 7-14
    *
    * @param trees
    * @return
    */
    public static TreeNode initTreeNode(Integer[] trees, int n) {
    if(n >= trees.length){
    return null;
    }
    if(null == trees[n]) {
    return null;
    }
    int l = n * 2 + 1;
    if(l > trees.length){
    return new TreeNode(trees[n],null,null);
    }
    TreeNode treeNode = new TreeNode(trees[n],initTreeNode(trees,2*n+1),initTreeNode(trees,2*n+2));
    return treeNode;
    }


    public static void main(String[] args) {
    Integer[] arr = {10, 5, -3, 3, 2, null, 11, 3, -2, null, 1};
    TreeNode treeNode = initTreeNode(arr, 0);
    }


    以前写的一个工具类
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