面试题 39，二叉树的深度，以及判断二叉树是否是平衡二叉树 (比较经典的题目)

比较老的题目，最简单版本的代码如下：

int CalcuDepth(TreeNode *root){
    if(NULL == root)
        return 0;
        
    int leftDepth = CalcuDepth(root -> m_pLeft);
    int rightDepth = CalcuDepth(root -> m_pRight);
    return (leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1);
}

 

接着还有第二问：判断二叉树是否是平衡树，即所有节点左右子树高度差不超过1。

先来个深搜再每次都调用上面的CalcuDepth函数的方法，肯定不能拿出来的。

说得过去的解法是需要把高度的计算和二叉平衡树的判断放在一起，从而避免重复遍历。

我的思路是，函数返回-1表明这个结点不是平衡二叉树，返回-1的条件要么就是左右孩子有一个返回了-1，要么就是当前结点不平衡。要是这些条件都不满足，还是照样返回较大深度。

看看根结点的返回值是不是-1就行了。这种思路比书上要简洁些。

代码：

#include <stdio.h>

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

int JudegeBalancedCore(BinaryTreeNode *root){
    if(NULL == root)
        return 0;
        
    int leftDepth = JudegeBalancedCore(root -> m_pLeft);
    int rightDepth = JudegeBalancedCore(root -> m_pRight);
    if(leftDepth == -1 || rightDepth == -1) return -1;
    if(leftDepth - rightDepth > 1 || rightDepth - leftDepth > 1) return -1;
    return (leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1);
}

//函数实现 
bool JudegeBalanced(BinaryTreeNode *root){
    if(NULL == root)
        return true;
    return JudegeBalancedCore(root) == -1 ? false : true;
}


// ====================树代码====================
BinaryTreeNode* CreateBinaryTreeNode(int value)
{
    BinaryTreeNode* pNode = new BinaryTreeNode();
    pNode->m_nValue = value;
    pNode->m_pLeft = NULL;
    pNode->m_pRight = NULL;

    return pNode;
}

void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
    if(pParent != NULL)
    {
        pParent->m_pLeft = pLeft;
        pParent->m_pRight = pRight;
    }
}

void PrintTreeNode(BinaryTreeNode* pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->m_nValue);

        if(pNode->m_pLeft != NULL)
            printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);
        else
            printf("left child is null.\n");

        if(pNode->m_pRight != NULL)
            printf("value of its right child is: %d.\n", pNode->m_pRight->m_nValue);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

    printf("\n");
}

void PrintTree(BinaryTreeNode* pRoot)
{
    PrintTreeNode(pRoot);

    if(pRoot != NULL)
    {
        if(pRoot->m_pLeft != NULL)
            PrintTree(pRoot->m_pLeft);

        if(pRoot->m_pRight != NULL)
            PrintTree(pRoot->m_pRight);
    }
}

void DestroyTree(BinaryTreeNode* pRoot)
{
    if(pRoot != NULL)
    {
        BinaryTreeNode* pLeft = pRoot->m_pLeft;
        BinaryTreeNode* pRight = pRoot->m_pRight;

        delete pRoot;
        pRoot = NULL;

        DestroyTree(pLeft);
        DestroyTree(pRight);
    }
}




// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot, bool expected)
{
    if(testName != NULL)
        printf("%s begins:\n", testName);

    if(JudegeBalanced(pRoot) == expected)
        printf("Passed.\n");
    else
        printf("Failed.\n");
}

// 完全二叉树
//             1
//         /      \
//        2        3
//       /\       / \
//      4  5     6   7
void Test1()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
    BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);

    ConnectTreeNodes(pNode1, pNode2, pNode3);
    ConnectTreeNodes(pNode2, pNode4, pNode5);
    ConnectTreeNodes(pNode3, pNode6, pNode7);

    Test("Test1", pNode1, true);

    DestroyTree(pNode1);
}

// 不是完全二叉树，但是平衡二叉树
//             1
//         /      \
//        2        3
//       /\         \
//      4  5         6
//        /
//       7
void Test2()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
    BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);

    ConnectTreeNodes(pNode1, pNode2, pNode3);
    ConnectTreeNodes(pNode2, pNode4, pNode5);
    ConnectTreeNodes(pNode3, NULL, pNode6);
    ConnectTreeNodes(pNode5, pNode7, NULL);

    Test("Test2", pNode1, true);

    DestroyTree(pNode1);
}

// 不是平衡二叉树
//             1
//         /      \
//        2        3
//       /\         
//      4  5        
//        /
//       6
void Test3()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);

    ConnectTreeNodes(pNode1, pNode2, pNode3);
    ConnectTreeNodes(pNode2, pNode4, pNode5);
    ConnectTreeNodes(pNode5, pNode6, NULL);

    Test("Test3", pNode1, false);

    DestroyTree(pNode1);
}


//               1
//              /
//             2
//            /
//           3
//          /
//         4
//        /
//       5
void Test4()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);

    ConnectTreeNodes(pNode1, pNode2, NULL);
    ConnectTreeNodes(pNode2, pNode3, NULL);
    ConnectTreeNodes(pNode3, pNode4, NULL);
    ConnectTreeNodes(pNode4, pNode5, NULL);

    Test("Test4", pNode1, false);

    DestroyTree(pNode1);
}

// 1
//  \
//   2
//    \
//     3
//      \
//       4
//        \
//         5
void Test5()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);

    ConnectTreeNodes(pNode1, NULL, pNode2);
    ConnectTreeNodes(pNode2, NULL, pNode3);
    ConnectTreeNodes(pNode3, NULL, pNode4);
    ConnectTreeNodes(pNode4, NULL, pNode5);

    Test("Test5", pNode1, false);

    DestroyTree(pNode1);
}

// 树中只有1个结点
void Test6()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    Test("Test6", pNode1, true);

    DestroyTree(pNode1);
}

// 树中没有结点
void Test7()
{
    Test("Test7", NULL, true);
}

int main()
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();
    Test7();

    return 0;
}

 

书上的思路是用一个引用的方式来额外存储路径值。相对麻烦一些。

代码：

bool IsBalanced(BinaryTreeNode* pRoot, int* pDepth)
{
    if(pRoot == NULL)
    {
        *pDepth = 0;
        return true;
    }

    int left, right;
    if(IsBalanced(pRoot->m_pLeft, &left) 
        && IsBalanced(pRoot->m_pRight, &right))
    {
        int diff = left - right;
        if(diff <= 1 && diff >= -1)
        {
            *pDepth = 1 + (left > right ? left : right);
            return true;
        }
    }

    return false;
}