剑指Offer面试题:17.树的子结构

一、题目:树的子结构

题目:输入两棵二叉树A和B,判断B是不是A的子结构。例如下图中的两棵二叉树,由于A中有一部分子树的结构和B是一样的,因此B是A的子结构。

  该二叉树的节点定义如下,这里使用C#语言描述:

    public class BinaryTreeNode
    {
        public int Data { get; set; }
        public BinaryTreeNode leftChild { get; set; }
        public BinaryTreeNode rightChild { get; set; }

        public BinaryTreeNode(int data)
        {
            this.Data = data;
        }

        public BinaryTreeNode(int data, BinaryTreeNode left, BinaryTreeNode right)
        {
            this.Data = data;
            this.leftChild = left;
            this.rightChild = right;
        }
    }

二、解题思路

2.1 核心步骤

  要查找树A中是否存在和树B结构一样的子树,我们可以分成两步:

  Step1.在树A中找到和B的根结点的值一样的结点R;

  Step2.判断树A中以R为根结点的子树是不是包含和树B一样的结构。

  很明显,这是一个递归的过程。

2.2 代码实现

    public static bool HasSubTree(BinaryTreeNode root1, BinaryTreeNode root2)
    {
        bool result = false;

        if (root1 != null && root2 != null)
        {
            if (root1.Data == root2.Data)
            {
                result = DoesTree1HasTree2(root1, root2);
            }
            // 从根节点的左子树开始匹配Tree2
            if (!result)
            {
                result = HasSubTree(root1.leftChild, root2);
            }
            // 如果左子树没有匹配成功则继续在右子树中继续匹配Tree2
            if (!result)
            {
                result = HasSubTree(root1.rightChild, root2);
            }
        }

        return result;
    }

    private static bool DoesTree1HasTree2(BinaryTreeNode root1, BinaryTreeNode root2)
    {
        if (root2 == null)
        {
            // 证明Tree2已经遍历结束,匹配成功
            return true;
        }

        if (root1 == null)
        {
            // 证明Tree1已经遍历结束,匹配失败
            return false;
        }

        if (root1.Data != root2.Data)
        {
            return false;
        }
        // 递归验证左子树和右子树是否包含Tree2
        return DoesTree1HasTree2(root1.leftChild, root2.leftChild) && DoesTree1HasTree2(root1.rightChild, root2.rightChild);
    }

三、单元测试

  为了方便测试,这里封装了一个设置指定根节点的左孩子和右孩子节点的方法:SetSubTreeNode

    public void SetSubTreeNode(BinaryTreeNode root, BinaryTreeNode lChild, BinaryTreeNode rChild)
    {
        if (root == null)
        {
            return;
        }

        root.leftChild = lChild;
        root.rightChild = rChild;
    }
View Code

3.1 功能测试

    // 01.树中结点含有分叉,树B是树A的子结构
    //                  8                8
    //              /       \           / \
    //             8         7         9   2
    //           /   \
    //          9     2
    //               / \
    //              4   7
    [TestMethod]
    public void HasSubTreeTest1()
    {
        BinaryTreeNode nodeA1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA2 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA3 = new BinaryTreeNode(7);
        BinaryTreeNode nodeA4 = new BinaryTreeNode(9);
        BinaryTreeNode nodeA5 = new BinaryTreeNode(2);
        BinaryTreeNode nodeA6 = new BinaryTreeNode(4);
        BinaryTreeNode nodeA7 = new BinaryTreeNode(7);

        SetSubTreeNode(nodeA1, nodeA2, nodeA3);
        SetSubTreeNode(nodeA2, nodeA4, nodeA5);
        SetSubTreeNode(nodeA5, nodeA6, nodeA7);

        BinaryTreeNode nodeB1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeB2 = new BinaryTreeNode(9);
        BinaryTreeNode nodeB3 = new BinaryTreeNode(2);

        SetSubTreeNode(nodeB1, nodeB2, nodeB3);

        Assert.AreEqual(SubTreeHelper.HasSubTree(nodeA1, nodeB1), true);
    }

    // 02.树中结点含有分叉,树B不是树A的子结构
    //                  8                8
    //              /       \           / \
    //             8         7         9   2
    //           /   \
    //          9     3
    //               / \
    //              4   7
    [TestMethod]
    public void HasSubTreeTest2()
    {
        BinaryTreeNode nodeA1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA2 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA3 = new BinaryTreeNode(7);
        BinaryTreeNode nodeA4 = new BinaryTreeNode(9);
        BinaryTreeNode nodeA5 = new BinaryTreeNode(3);
        BinaryTreeNode nodeA6 = new BinaryTreeNode(4);
        BinaryTreeNode nodeA7 = new BinaryTreeNode(7);

        SetSubTreeNode(nodeA1, nodeA2, nodeA3);
        SetSubTreeNode(nodeA2, nodeA4, nodeA5);
        SetSubTreeNode(nodeA5, nodeA6, nodeA7);

        BinaryTreeNode nodeB1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeB2 = new BinaryTreeNode(9);
        BinaryTreeNode nodeB3 = new BinaryTreeNode(2);

        SetSubTreeNode(nodeB1, nodeB2, nodeB3);

        Assert.AreEqual(SubTreeHelper.HasSubTree(nodeA1, nodeB1), false);
    }

3.2 特殊输入测试

    // 03.树中结点只有左子结点,树B是树A的子结构
    //                8                  8
    //              /                   / 
    //             8                   9   
    //           /                    /
    //          9                    2
    //         /      
    //        2        
    //       /
    //      5
    [TestMethod]
    public void HasSubTreeTest3()
    {
        BinaryTreeNode nodeA1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA2 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA3 = new BinaryTreeNode(9);
        BinaryTreeNode nodeA4 = new BinaryTreeNode(2);
        BinaryTreeNode nodeA5 = new BinaryTreeNode(5);

        nodeA1.leftChild = nodeA2;
        nodeA2.leftChild = nodeA3;
        nodeA3.leftChild = nodeA4;
        nodeA4.leftChild = nodeA5;

        BinaryTreeNode nodeB1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeB2 = new BinaryTreeNode(9);
        BinaryTreeNode nodeB3 = new BinaryTreeNode(2);

        nodeB1.leftChild = nodeB2;
        nodeB2.leftChild = nodeB3;

        Assert.AreEqual(SubTreeHelper.HasSubTree(nodeA1, nodeB1), true);
    }

    // 04.树中结点只有左子结点,树B不是树A的子结构
    //                8                  8
    //              /                   / 
    //             8                   9   
    //           /                    /
    //          9                    3
    //         /      
    //        2        
    //       /
    //      5
    [TestMethod]
    public void HasSubTreeTest4()
    {
        BinaryTreeNode nodeA1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA2 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA3 = new BinaryTreeNode(9);
        BinaryTreeNode nodeA4 = new BinaryTreeNode(2);
        BinaryTreeNode nodeA5 = new BinaryTreeNode(5);

        nodeA1.leftChild = nodeA2;
        nodeA2.leftChild = nodeA3;
        nodeA3.leftChild = nodeA4;
        nodeA4.leftChild = nodeA5;

        BinaryTreeNode nodeB1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeB2 = new BinaryTreeNode(9);
        BinaryTreeNode nodeB3 = new BinaryTreeNode(3);

        nodeB1.leftChild = nodeB2;
        nodeB2.leftChild = nodeB3;

        Assert.AreEqual(SubTreeHelper.HasSubTree(nodeA1, nodeB1), false);
    }

    // 05.树中结点只有右子结点,树B是树A的子结构
    //       8                   8
    //        \                   \ 
    //         8                   9   
    //          \                   \
    //           9                   2
    //            \      
    //             2        
    //              \
    //               5
    [TestMethod]
    public void HasSubTreeTest5()
    {
        BinaryTreeNode nodeA1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA2 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA3 = new BinaryTreeNode(9);
        BinaryTreeNode nodeA4 = new BinaryTreeNode(2);
        BinaryTreeNode nodeA5 = new BinaryTreeNode(5);

        nodeA1.rightChild = nodeA2;
        nodeA2.rightChild = nodeA3;
        nodeA3.rightChild = nodeA4;
        nodeA4.rightChild = nodeA5;

        BinaryTreeNode nodeB1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeB2 = new BinaryTreeNode(9);
        BinaryTreeNode nodeB3 = new BinaryTreeNode(2);

        nodeB1.rightChild = nodeB2;
        nodeB2.rightChild = nodeB3;

        Assert.AreEqual(SubTreeHelper.HasSubTree(nodeA1, nodeB1), true);
    }

    // 06.树中结点只有右子结点,树B不是树A的子结构
    //       8                   8
    //        \                   \ 
    //         8                   9   
    //          \                 / \
    //           9               3   2
    //            \      
    //             2        
    //              \
    //               5
    [TestMethod]
    public void HasSubTreeTest6()
    {
        BinaryTreeNode nodeA1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA2 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA3 = new BinaryTreeNode(9);
        BinaryTreeNode nodeA4 = new BinaryTreeNode(2);
        BinaryTreeNode nodeA5 = new BinaryTreeNode(5);

        nodeA1.rightChild = nodeA2;
        nodeA2.rightChild = nodeA3;
        nodeA3.rightChild = nodeA4;
        nodeA4.rightChild = nodeA5;

        BinaryTreeNode nodeB1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeB2 = new BinaryTreeNode(9);
        BinaryTreeNode nodeB3 = new BinaryTreeNode(3);
        BinaryTreeNode nodeB4 = new BinaryTreeNode(2);

        nodeB1.rightChild = nodeB2;
        SetSubTreeNode(nodeB2, nodeB3, nodeB4);

        Assert.AreEqual(SubTreeHelper.HasSubTree(nodeA1, nodeB1), false);
    }

    // 07.树A为空树
    [TestMethod]
    public void HasSubTreeTest7()
    {
        BinaryTreeNode nodeB1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeB2 = new BinaryTreeNode(9);
        BinaryTreeNode nodeB3 = new BinaryTreeNode(3);
        BinaryTreeNode nodeB4 = new BinaryTreeNode(2);

        nodeB1.rightChild = nodeB2;
        SetSubTreeNode(nodeB2, nodeB3, nodeB4);

        Assert.AreEqual(SubTreeHelper.HasSubTree(null, nodeB1), false);
    }

    // 08.树B为空树
    [TestMethod]
    public void HasSubTreeTest8()
    {
        BinaryTreeNode nodeA1 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA2 = new BinaryTreeNode(8);
        BinaryTreeNode nodeA3 = new BinaryTreeNode(9);
        BinaryTreeNode nodeA4 = new BinaryTreeNode(2);
        BinaryTreeNode nodeA5 = new BinaryTreeNode(5);

        nodeA1.rightChild = nodeA2;
        nodeA2.rightChild = nodeA3;
        nodeA3.rightChild = nodeA4;
        nodeA4.rightChild = nodeA5;

        Assert.AreEqual(SubTreeHelper.HasSubTree(nodeA1, null), false);
    }

    // 09.树A和树B都为空树
    [TestMethod]
    public void HasSubTreeTest9()
    {
        Assert.AreEqual(SubTreeHelper.HasSubTree(null, null), false);
    }

3.3 测试结果

  (1)测试通过情况

  (2)代码覆盖率

 

posted @ 2015-08-30 23:05 Edison Chou 阅读(...) 评论(...) 编辑 收藏