题目：输入一棵二叉树的根结点，判断该树是不是平衡二叉树

题目：输入一棵二叉树的根结点，判断该树是不是平衡二叉树。如果某二叉树中任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。例如，下图中的二叉树就是一棵平衡二叉树。

 


解题思路：

（1）需要重复遍历节点多次的解法

 

      在上一篇博客中（http://blog.csdn.net/yanxiaolx/article/details/52282776），有了求二叉树的深度的经验之后再解决这个问题，我们很容易就能想到一个思路：在遍历树的每个结点的时候，调用函数TreeDepth得到它的左右子树的深度。如果每个结点的左右子树的深度相差都不超过1，按照定义它就是一棵平衡的二叉树。

代码实现：

// ====================方法1====================
//求二叉树的深度
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);
}

//需要重复遍历节点的算法，不够好
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);
}

       上面的代码固然简洁，但我们也要注意到由于一个结点会被重复遍历多次，这种思路的时间效率不高。例如在IsBalancedBinaryTree方法中输入上图中的二叉树，我们将首先判断根结点（结点1）是不是平衡的。此时我们往函数TreeDepth输入左子树的根结点（结点2）时，需要遍历结点4、5、7。接下来判断以结点2为根结点的子树是不是平衡树的时候，仍然会遍历结点4、5、7。毫无疑问，重复遍历同一个结点会影响性能。

 

（2）每个节点只需遍历一次的解法

 

　　换个角度来思考，如果我们用后序遍历的方式遍历二叉树的每一个结点，在遍历到一个结点之前我们就已经遍历了它的左右子树。只要在遍历每个结点的时候记录它的深度（某一结点的深度等于它到叶节点的路径的长度），我们就可以一边遍历一边判断每个结点是不是平衡的。

代码实现：

// ====================方法2====================
//后序每个节点只遍历一次，一边遍历一边记录它的深度
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;
}

bool  IsBalanced_Solution2(BinaryTreeNode* pRoot)
{
int depth=0;
return IsBalanced(pRoot, &depth);
}

 

       在上面的代码中，我们用后序遍历的方式遍历整棵二叉树。在遍历某结点的左右子结点之后，我们可以根据它的左右子结点的深度判断它是不是平衡的，并得到当前结点的深度。当最后遍历到树的根结点的时候，也就判断了整棵二叉树是不是平衡二叉树。

 

完整代码及其测试用例实现：

 

#include<iostream>
using namespace std;

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;

return pNode;
}

void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
if (pParent != NULL)
{
pParent->m_pLeft = pLeft;
pParent->m_pRight = 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====================
//求二叉树的深度
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);
}

//需要重复遍历节点的算法，不够好
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====================
//后序每个节点只遍历一次，一边遍历一边记录它的深度
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;
}

bool  IsBalanced_Solution2(BinaryTreeNode* pRoot)
{
int depth=0;
return IsBalanced(pRoot, &depth);
}

// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot, bool expected)
{
if (testName != NULL)
{
cout << testName<< " begins:" << endl;
}

cout << "Solution1 begins: " ;

if (IsBalanced_Solution1(pRoot) == expected)
{
cout << "Passed." << endl;
}
else
{
cout << "Failed." << endl;
}

cout << "Solution2 begins: ";

if (IsBalanced_Solution2(pRoot) == expected)
{
cout << "Passed." << endl;
}
else
{
cout << "Failed." << endl;
}
}

void Test1()
{
// 完全二叉树
//             1
//         /      \
//        2        3
//       /\       / \
//      4  5     6   7
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);
}

void Test2()
{
// 不是完全二叉树，但是平衡二叉树
//             1
//         /      \
//        2        3
//       /\         \
//      4  5         6
//        /
//       7
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);
}

void Test3()
{
// 不是平衡二叉树
//             1
//         /      \
//        2        3
//       /\
//      4  5
//        /
//       6
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);
}

void Test4()
{
//               1
//              /
//             2
//            /
//           3
//          /
//         4
//        /
//       5
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);
}

void Test5()
{
// 1
//  \
//   2
//    \
//     3
//      \
//       4
//        \
//         5
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);
}

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

DestroyTree(pNode1);
}

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

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

system("pause");
return 0;
}

运行结果：

Test1 begins:

Solution1 begins: Passed.

Solution2 begins: Passed.

Test2 begins:

Solution1 begins: Passed.

Solution2 begins: Passed.

Test3 begins:

Solution1 begins: Passed.

Solution2 begins: Passed.

Test4 begins:

Solution1 begins: Passed.

Solution2 begins: Passed.

Test5 begins:

Solution1 begins: Passed.

Solution2 begins: Passed.

Test6 begins:

Solution1 begins: Passed.

Solution2 begins: Passed.

Test7 begins:

Solution1 begins: Passed.

Solution2 begins: Passed.

请按任意键继续. . .