Question

我早就被要求撰写一种复习方法,以调查是否有单身儿童。我收到了一些基本案件,但是,我对如何处理休养科感到困惑不解,因为如果其中一名儿童,如果他们中有0名儿童或再次生育,我就需要调查权利,然后调查遗留的子树,并作假。

我迄今所做的是:

public static boolean noSingleChildren( BinaryTreeNode t ) { 
    if (rightC == null || leftC == null) {
         return false;
    } else if (rightC == null && leftC == null) {
        return true;
    } else {
        return............
    }
}

Answer 1

逻辑非常简单:

If the current node only has a single child, you re done.
Otherwise, recursively ask each non-null child the same question, and combine the answers using logical "or".

既然看上去做家务,我就把工作交给你。

Answer 2

public static boolean noSingleChildren( BinaryTreeNode t ) { 
    if (rightC == null || leftC == null) {
         return false;
    } else if (rightC == null && leftC == null) {
        return true;
    } else {
        return noSingleChildren(t.getLeftBranch()) || noSingleChildren(t.getRightBranch());
    }
}

Answer 3

Ho,我爱树:

public static boolean hasSingleChildren( BinaryTreeNode t ) { 
    if (t == null) {
         return false;
    } else if (t.rightC == null && t.leftC != null) {
        return true;
    } else if (t.rightC != null && t.leftC == null) {
        return true;
    } else {
        return hasSingleChildren(t.rightC) || hasSingleChildren(t.leftC);
    }
}

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