题目:输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
思路:
先序遍历的第一个元素为根节点,在中序遍历中找到这个根节点,从而可以将中序遍历分为左右两个部分,左边部分为左子树的中序遍历,右边部分为右子树的中序遍历,进而也可以将先序遍历除第一个元素以外的剩余部分分为两个部分,第一个部分为左子树的先序遍历,第二个部分为右子树的先序遍历。
由上述分析结果,可以递归调用构建函数,根据左子树、右子树的先序、中序遍历重建左、右子树。
完整代码及其测试用例实现:
#include<iostream>
#include <exception>//异常处理类
using namespace std;
//二叉树节点定义
struct BinaryTreeNode
{
int m_nValue;
BinaryTreeNode* m_pLeft;
BinaryTreeNode* m_pRight;
};
//打印树节点
void PrintTreeNode(BinaryTreeNode* pNode)
{
if (pNode != NULL)
{
cout << "value of this root node is: " << pNode->m_nValue << endl;
if (pNode->m_pLeft != NULL)
{
cout << "value of its left child is: " << pNode->m_pLeft->m_nValue << endl;
}
else
{
cout << "left child is null. " << endl;
}
if (pNode->m_pRight != NULL)
{
cout << "value of its right child is: " << pNode->m_pRight->m_nValue << endl;
}
else
{
cout << "right child is null. " << endl;
}
}
else
{
cout << "this root node is null. " << endl;
}
cout << endl;
}
//遍历树
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);
}
}
}
//销毁树
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);
}
}
//重建二叉树
BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder,int* startInorder, int* endInorder)
{
// 前序遍历序列的第一个数字是根结点的值
int rootValue = startPreorder[0];
BinaryTreeNode* root = new BinaryTreeNode();
root->m_nValue = rootValue;
root->m_pLeft = NULL;
root->m_pRight = NULL;
if (startPreorder == endPreorder)
{
if (startInorder == endInorder && *startPreorder == *startInorder)
{
return root;
}
else
{
throw std::exception("Invalid input.");
}
}
// 在中序遍历中找到根结点的值
int* rootInorder = startInorder;
while (rootInorder <= endInorder && *rootInorder != rootValue)
{
++rootInorder;
}
if (rootInorder == endInorder && *rootInorder != rootValue)
{
throw std::exception("Invalid input.");
}
int leftLength = rootInorder - startInorder;
int* leftPreorderEnd = startPreorder + leftLength;
if (leftLength > 0)
{
// 构建左子树
root->m_pLeft = ConstructCore(startPreorder + 1, leftPreorderEnd,
startInorder, rootInorder - 1);
}
if (leftLength < endPreorder - startPreorder)
{
// 构建右子树
root->m_pRight = ConstructCore(leftPreorderEnd + 1, endPreorder,
rootInorder + 1, endInorder);
}
return root;
}
BinaryTreeNode* Construct(int* preorder, int* inorder, int length)
{
if (preorder == NULL || inorder == NULL || length <= 0)
{
return NULL;
}
return ConstructCore(preorder, preorder + length - 1,
inorder, inorder + length - 1);
}
// ====================测试代码====================
void Test(char* testName, int* preorder, int* inorder, int length)
{
if (testName != NULL)
{
cout << testName << " begins:" << endl;
}
cout << "The preorder sequence is:" ;
for (int i = 0; i < length; ++i)
{
cout << preorder[i] << " ";
}
cout <<endl;
cout << "The inorder sequence is:";
for (int i = 0; i < length; ++i)
{
cout << inorder[i] << " ";
}
cout << endl;
try
{
BinaryTreeNode* root = Construct(preorder, inorder, length);
PrintTree(root);
DestroyTree(root);
}
catch (std::exception& exception)
{
cout << " Invalid Input."<<endl;
}
}
void Test1()
{
// 普通二叉树
// 1
// / \
// 2 3
// / / \
// 4 5 6
// \ /
// 7 8
const int length = 8;
int preorder[length] = { 1, 2, 4, 7, 3, 5, 6, 8 };
int inorder[length] = { 4, 7, 2, 1, 5, 3, 8, 6 };
Test("Test1", preorder, inorder, length);
}
void Test2()
{
// 所有结点都没有右子结点
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
const int length = 5;
int preorder[length] = { 1, 2, 3, 4, 5 };
int inorder[length] = { 5, 4, 3, 2, 1 };
Test("Test2", preorder, inorder, length);
}
void Test3()
{
// 所有结点都没有左子结点
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
const int length = 5;
int preorder[length] = { 1, 2, 3, 4, 5 };
int inorder[length] = { 1, 2, 3, 4, 5 };
Test("Test3", preorder, inorder, length);
}
void Test4()
{
// 树中只有一个结点
const int length = 1;
int preorder[length] = { 1 };
int inorder[length] = { 1 };
Test("Test4", preorder, inorder, length);
}
void Test5()
{
// 完全二叉树
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
const int length = 7;
int preorder[length] = { 1, 2, 4, 5, 3, 6, 7 };
int inorder[length] = { 4, 2, 5, 1, 6, 3, 7 };
Test("Test5", preorder, inorder, length);
}
void Test6()
{
// 输入空指针
Test("Test6", NULL, NULL, 0);
}
void Test7()
{
// 输入的两个序列不匹配
const int length = 7;
int preorder[length] = { 1, 2, 4, 5, 3, 6, 7 };
int inorder[length] = { 4, 2, 8, 1, 6, 3, 7 };
Test("Test7: for unmatched input", preorder, inorder, length);
}
int main()
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
system("pause");
return 0;
}
运行结果:
Test1 begins:
The preorder sequence is:1 2 4 7 3 5 6 8
The inorder sequence is:4 7 2 1 5 3 8 6
value of this root node is: 1
value of its left child is: 2
value of its right child is: 3
value of this root node is: 2
value of its left child is: 4
right child is null.
value of this root node is: 4
left child is null.
value of its right child is: 7
value of this root node is: 7
left child is null.
right child is null.
value of this root node is: 3
value of its left child is: 5
value of its right child is: 6
value of this root node is: 5
left child is null.
right child is null.
value of this root node is: 6
value of its left child is: 8
right child is null.
value of this root node is: 8
left child is null.
right child is null.
Test2 begins:
The preorder sequence is:1 2 3 4 5
The inorder sequence is:5 4 3 2 1
value of this root node is: 1
value of its left child is: 2
right child is null.
value of this root node is: 2
value of its left child is: 3
right child is null.
value of this root node is: 3
value of its left child is: 4
right child is null.
value of this root node is: 4
value of its left child is: 5
right child is null.
value of this root node is: 5
left child is null.
right child is null.
Test3 begins:
The preorder sequence is:1 2 3 4 5
The inorder sequence is:1 2 3 4 5
value of this root node is: 1
left child is null.
value of its right child is: 2
value of this root node is: 2
left child is null.
value of its right child is: 3
value of this root node is: 3
left child is null.
value of its right child is: 4
value of this root node is: 4
left child is null.
value of its right child is: 5
value of this root node is: 5
left child is null.
right child is null.
Test4 begins:
The preorder sequence is:1
The inorder sequence is:1
value of this root node is: 1
left child is null.
right child is null.
Test5 begins:
The preorder sequence is:1 2 4 5 3 6 7
The inorder sequence is:4 2 5 1 6 3 7
value of this root node is: 1
value of its left child is: 2
value of its right child is: 3
value of this root node is: 2
value of its left child is: 4
value of its right child is: 5
value of this root node is: 4
left child is null.
right child is null.
value of this root node is: 5
left child is null.
right child is null.
value of this root node is: 3
value of its left child is: 6
value of its right child is: 7
value of this root node is: 6
left child is null.
right child is null.
value of this root node is: 7
left child is null.
right child is null.
Test6 begins:
The preorder sequence is:
The inorder sequence is:
this root node is null.
Test7: for unmatched input begins:
The preorder sequence is:1 2 4 5 3 6 7
The inorder sequence is:4 2 8 1 6 3 7
Invalid Input.
请按任意键继续. . .