二叉搜索树的第k个节点

问题:给定一颗二叉搜索树，请找出其中的第k大的结点。例如， 5 / \ 3 7 /\ /\ 2 4 6 8 中，按结点数值大小顺序第三个结点的值为4。



  分析：因为是二叉搜索树，所以可以用中序遍历的方式存储然后输出第k个结点，要注意为0和大于结点大小的K值；

详细的设计代码如下:

//二叉搜索树的第k个节点
BinaryTreeNode* KthNode(BinaryTreeNode* pRoot, unsigned int k)
{
if(pRoot == NULL || k == 0)
return NULL;

return KthNodeCore(pRoot, k);
}

BinaryTreeNode* KthNodeCore(BinaryTreeNode* pRoot, unsigned int& k)
{
BinaryTreeNode* target = NULL;

if(pRoot->m_pLeft != NULL)
target = KthNodeCore(pRoot->m_pLeft, k);

if(target == NULL)
{
if(k == 1)
target = pRoot;

k--;
}

if(target == NULL && pRoot->m_pRight != NULL)
target = KthNodeCore(pRoot->m_pRight, k);

return target;
}

void Test(char* testName, BinaryTreeNode* pRoot, unsigned int k, bool isNull, int expected)
{
if(testName != NULL)
printf("%s begins: ", testName);

BinaryTreeNode* pTarget = KthNode(pRoot, k);
if((isNull && pTarget == NULL) || (!isNull && pTarget->m_nValue == expected))
printf("Passed.\n");
else
printf("FAILED.\n");
}

// 8
// 6 10
// 5 7 9 11
void TestA()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);

ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);

Test("TestA0", pNode8, 0, true, -1);
Test("TestA1", pNode8, 1, false, 5);
Test("TestA2", pNode8, 2, false, 6);
Test("TestA3", pNode8, 3, false, 7);
Test("TestA4", pNode8, 4, false, 8);
Test("TestA5", pNode8, 5, false, 9);
Test("TestA6", pNode8, 6, false, 10);
Test("TestA7", pNode8, 7, false, 11);
Test("TestA8", pNode8, 8, true, -1);

DestroyTree(pNode8);

printf("\n\n");
}

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

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

Test("TestB0", pNode5, 0, true, -1);
Test("TestB1", pNode5, 1, false, 1);
Test("TestB2", pNode5, 2, false, 2);
Test("TestB3", pNode5, 3, false, 3);
Test("TestB4", pNode5, 4, false, 4);
Test("TestB5", pNode5, 5, false, 5);
Test("TestB6", pNode5, 6, true, -1);

DestroyTree(pNode5);

printf("\n\n");
}

// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void TestC()
{
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("TestC0", pNode1, 0, true, -1);
Test("TestC1", pNode1, 1, false, 1);
Test("TestC2", pNode1, 2, false, 2);
Test("TestC3", pNode1, 3, false, 3);
Test("TestC4", pNode1, 4, false, 4);
Test("TestC5", pNode1, 5, false, 5);
Test("TestC6", pNode1, 6, true, -1);

DestroyTree(pNode1);

printf("\n\n");
}

// There is only one node in a tree
void TestD()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);

Test("TestD0", pNode1, 0, true, -1);
Test("TestD1", pNode1, 1, false, 1);
Test("TestD2", pNode1, 2, true, -1);

DestroyTree(pNode1);

printf("\n\n");
}

// empty tree
void TestE()
{
Test("TestE0", NULL, 0, true, -1);
Test("TestE1", NULL, 1, true, -1);

printf("\n\n");
}

int main(int argc, char* argv[])
{
TestA();
TestB();
TestC();
TestD();
TestE();
}