如何在树中搜索节点并返回它？ [英] How to search for a node in a tree and return it?

问题描述

我正在尝试在二叉树中搜索节点并在其存在的情况下返回，否则返回null。顺便说一下，节点类有一个方法名（）返回一个带有它名字的字符串......到目前为止我所拥有的是：

I'm trying to search for a node in a binary tree and return in case it's there, otherwise, return null. By the way, the node class has a method name() that return a string with it's name...What I have so far is:

private Node search(String name, Node node){

     if(node != null){
         if(node.name().equals(name)){
            return node;
         }

      else{
         search(name, node.left);
         search(name, node.right);
      }
    }
    return null;
}

这是正确的吗？

推荐答案

如果结果不为null，您需要确保对搜索的递归调用返回。

You need to make sure your recursive calls to search return if the result isn't null.

类似于这应该工作...

Something like this should work...

private Node search(String name, Node node){
    if(node != null){
        if(node.name().equals(name)){
           return node;
        } else {
            Node foundNode = search(name, node.left);
            if(foundNode == null) {
                foundNode = search(name, node.right);
            }
            return foundNode;
         }
    } else {
        return null;
    }
}

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