題目:判斷一個二叉排序樹是否是平衡二叉樹
思路:利用遞歸判斷左右子樹的深度是否相差1來判斷是否是平衡二叉樹。
#include<stdio.h>
#include "stdafx.h"
struct BinaryTreeNode
{
int m_nValue;
BinaryTreeNode* m_pLeft;
BinaryTreeNode* m_pRight;
};
BinaryTreeNode* CreateBinaryTreeNode(int value)
{
BinaryTreeNode* pNode = new BinaryTreeNode();
pNode->m_nValue = value;
pNode->m_pLeft = NULL;
pNode->m_pRight = NULL;
}
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);
}
}
//========================方法1==============================
int TreeDepth(BinaryTreeNode* pRoot)
{
if(pRoot == NULL)
return 0;
int nLeft = TreeDepth(pRoot->m_pLeft);
int nRight = TreeDepth(pRoot->m_pRight);
return (nLeft > nRight) ? (nLeft + 1) : (nRight + 1);
}
bool IsBalanced_Solution1(BinaryTreeNode* pRoot)
{
if(pRoot == NULL)
return true;
int left = TreeDepth(pRoot->m_pLeft);
int right = TreeDepth(pRoot->m_pRight);
int diff = left - right;
if(diff > 1 || diff < -1)
return false;
return IsBalanced_Solution1(pRoot->m_pLeft)
&& IsBalanced_Solution1(pRoot->m_pRight);
}
//=====================方法2===========================
bool IsBalanced(BinaryTreeNode* pRoot, int* pDepth);
bool IsBalanced_Solution2(BinaryTreeNode* pRoot)
{
int depth = 0;
return IsBalanced(pRoot, &depth);
}
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;
}
// 不是完全二叉樹,但是平衡二叉樹
// 1
// / \
// 2 3
// /\ \
// 4 5 6
// /
// 7
int main()
{
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);
printf("Solution1 begins: ");
if(IsBalanced_Solution1(pNode1))
printf("is balanced.\n");
else
printf("not balanced.\n");
printf("Solution2 begins: ");
if(IsBalanced_Solution2(pNode1))
printf("is balanced.\n");
else
printf("not balanced.\n");
printf("\n");
DestroyTree(pNode1);
return 0;
}
二叉樹的常見問題及其解決程序 http://www.linuxidc.com/Linux/2013-04/83661.htm
【遞歸】二叉樹的先序建立及遍歷 http://www.linuxidc.com/Linux/2012-12/75608.htm
在JAVA中實現的二叉樹結構 http://www.linuxidc.com/Linux/2008-12/17690.htm
【非遞歸】二叉樹的建立及遍歷 http://www.linuxidc.com/Linux/2012-12/75607.htm
二叉樹遞歸實現與二重指針 http://www.linuxidc.com/Linux/2013-07/87373.htm
二叉樹先序中序非遞歸算法 http://www.linuxidc.com/Linux/2014-06/102935.htm
輕松搞定面試中的二叉樹題目 http://www.linuxidc.com/linux/2014-07/104857.htm