递归搜索非二叉树中的节点 [英] Recursive search for a node in non-binary tree

问题描述

我想在非二叉树中搜索一个项目（任何节点都可以有n个子节点）并立即退出递归。有问题的节点可以是任何节点，而不仅仅是叶子。

I want to search for an item in non-binary tree (any node can have n - children) and exit from recursion immediately. The node in question can be any node, not only leafs.

这是我的代码，但我没有完全搜索。

This is my code but i don't get complete search.

private nNode recursiveSearch(data gi,nNode node){
        if (node.getdata()==gi)
            return node;
        nNode[] children = node.getChildren(); 
        if (children.length>0)
        for (int i = 0; i < children.length; i++) {         
            return recursiveSearch(gi, children[i]);
        }
        return null;
 }

nNode包含：

ArrayList mChildren; （它是孩子们）

和数据对象。

ArrayList mChildren ; (it's children)
and data object.

推荐答案

探索第一个孩子后不应退出。对于循环，前面的 if 语句不需要。

You shouldn't exit after exploring the first child. You don't need the if statement in front of the for loop.

private nNode recursiveSearch(data gi,nNode node){
    if (node.getdata()==gi)
        return node;
    nNode[] children = node.getChildren(); 
    nNode res = null;
    for (int i = 0; res == null && i < children.length; i++) {         
        res = recursiveSearch(gi, children[i]);
    }
    return res;
 }

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