树的子结构

2024年5月5日 140次阅读

    输入两棵二叉树A和B，判断B是不是A的子结构

    C#实现：

#region 树的子结构
        /// 输入两棵二叉树A和B，判断B是不是A的子结构
        /// 
        public static bool HasSubTree(BinaryTreeNode pRoot1, BinaryTreeNode pRoot2)
        {
            bool result = false;

            if (pRoot1 != null && pRoot2 != null)
            {
                if (pRoot1.value == pRoot2.value)
                    result = DoesTree1HaveTree2(pRoot1, pRoot2);
                if (!result)
                    result = HasSubTree(pRoot1.left, pRoot2);
                if (!result)
                    result = HasSubTree(pRoot1.right, pRoot2);
            }

            return result;
        }

        public static bool DoesTree1HaveTree2(BinaryTreeNode pRoot1, BinaryTreeNode pRoot2)
        {
            if (pRoot2 == null)
                return true;
            if (pRoot1 == null)
                return false;
            if (pRoot1.value != pRoot2.value)
                return false;

            return DoesTree1HaveTree2(pRoot1.left, pRoot2.left) &&
                DoesTree1HaveTree2(pRoot1.right, pRoot2.right);
        }
        #endregion

    Java实现：

/**
	 * 树的子结构 输入两棵二叉树A和B，判断B是不是A的子结构
	 * 
	 * @param pRoot1
	 * @param pRoot2
	 * @return
	 */
	public static boolean hasSubTree(BinaryTreeNode pRoot1, BinaryTreeNode pRoot2) {
		boolean result = false;

		if (pRoot1 != null && pRoot2 != null) {
			if (pRoot1.value == pRoot2.value)
				result = doesTree1HaveTree2(pRoot1, pRoot2);
			if (!result)
				result = hasSubTree(pRoot1.left, pRoot2);
			if (!result)
				result = hasSubTree(pRoot1.right, pRoot2);
		}

		return result;
	}

	private static boolean doesTree1HaveTree2(BinaryTreeNode pRoot1, BinaryTreeNode pRoot2) {
		if (pRoot2 == null)
			return true;
		if (pRoot1 == null)
			return false;
		if (pRoot1.value != pRoot2.value)
			return false;
		return doesTree1HaveTree2(pRoot1.left, pRoot2.left) && doesTree1HaveTree2(pRoot1.right, pRoot2.right);
	}

    Python实现：

@staticmethod
    def hasSubTree(pRoot1, pRoot2):
        """
        树的子结构 输入两棵二叉树A和B，判断B是不是A的子结构
        :param pRoot1: 
        :param pRoot2: 
        :return: 
        """
        result = False
        
        if pRoot1 != None and pRoot2 != None:
            if pRoot1.value == pRoot2.value:
                result = BinaryTree.doesTree1HaveTree2(pRoot1, pRoot2)
            if not result:
                result = BinaryTree.hasSubTree(pRoot1.left, pRoot2)
            if not result:
                result = BinaryTree.hasSubTree(pRoot1.right, pRoot2)
                
        return result
    
    @staticmethod
    def doesTree1HaveTree2(pRoot1, pRoot2):
        if pRoot2 == None:
            return True
        if pRoot1 == None:
            return False
        if pRoot1.value != pRoot2.value:
            return False
        return BinaryTree.doesTree1HaveTree2(pRoot1.left, 
                                             pRoot2.left) \
               and BinaryTree.doesTree1HaveTree2(pRoot1.right, 
                                                 pRoot2.right)

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