18. 樹的子結構

題目描述

輸入兩顆二叉樹A，B，判斷B是不是A的子結構。

struct BinaryTreeNode 
{
    int                    m_nValue; 
    BinaryTreeNode*        m_pLeft;  
    BinaryTreeNode*        m_pRight; 
};

解析

遞歸調用HasSubtree遍歷二叉樹A，如果發現某一結點的值與樹B的頭結點的值相同，則調用DoesTree1HaveTree2，判斷以這兩個相等結點爲根結點的樹，是否相等（根結點值相等，左子樹相等，右子樹相等）

實現

bool HasSubtree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2){
    bool result = false;

    if (pRoot1 != NULL && pRoot2 != NULL){
        if (pRoot1->m_nValue == pRoot2->m_nValue)
            result = DoesTree1HaveTree2(pRoot1, pRoot2);
        if (!result)
            result = HasSubtree(pRoot1->m_pLeft, pRoot2);
        if (!result)
            result = HasSubtree(pRoot1->m_pRight, pRoot2);
    }
    return result;
}
bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2){
    if (pRoot2 == NULL)
        return true;
    if (pRoot1 == NULL)
        return false;
    if (pRoot1->m_nValue != pRoot2->m_nValue)
        return false;

    return DoesTree1HaveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) &&
        DoesTree1HaveTree2(pRoot1->m_pRight, pRoot2->m_pRight);
}