面试题39：平衡二叉树判断

1.输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中的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。

分析：

首先要的获得二叉树的深度信息，很明显，当二叉树只有一个结点时，深度为1，当二叉树只有左子树时，深度为左子树的深度+1，当只有右子树时，高度为右子树的高度+1，当左右子树都有时，为左右子树中深度的较大值+1.可以采用递归的方式实现。这样可以获得二叉树的深度信息。

然后，再判断二叉树是否满足平衡二叉树的条件。前面已经计算得到树的深度的信息，基于此，可以对树的每个结点的左右子树的深度信息是否满足条件，但是这种方法存在着大量的重复的计算。如果使用后序遍历的方式遍历二叉树的每个结点，在每遍历到一个结点之前我们就已经遍历了它的左右子树。这样在遍历每个结点的时候记录它的深度，就可以一边遍历一边判断是否满足条件。

源码：

/*** 功能说明：Description* 作者：K0713* 日期：2016-9-12**/#include<iostream>using namespace std;//二叉树结构struct BinaryTreeNode{int                    m_nValue;BinaryTreeNode*        m_pLeft;BinaryTreeNode*        m_pRight;};//创建二叉树的结点BinaryTreeNode* CreateBinaryTreeNode(int value);//连接二叉树的结点void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight);//打印二叉树的结点void PrintTreeNode(BinaryTreeNode* pNode);//先序遍历二叉树void PrintTree(BinaryTreeNode* pRoot);//二叉树销毁void DestroyTree(BinaryTreeNode* pRoot);//树的深度信息int TreeDepth(BinaryTreeNode* pRoot);//逐个判断每个结点的左右子树深度bool IsBalanced_Solution1(BinaryTreeNode* pRoot);bool IsBalanced(BinaryTreeNode* pRoot, int* pDepth);bool IsBalanced_Solution2(BinaryTreeNode* pRoot);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, NULL, pNode6);ConnectTreeNodes(pNode5, pNode7, NULL);bool result=IsBalanced_Solution2(pNode1);cout << "the result is: ";if (result == true)cout << "balance" << endl;elsecout << "not balance" << endl;system("PAUSE");return 0;}//创建树结点BinaryTreeNode* CreateBinaryTreeNode(int value){BinaryTreeNode* pNode = new BinaryTreeNode();pNode->m_nValue = value;pNode->m_pLeft = NULL;pNode->m_pRight = NULL;return pNode;}//连接树结点void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight){if (pParent != NULL){pParent->m_pLeft = pLeft;pParent->m_pRight = pRight;}}//打印void PrintTreeNode(BinaryTreeNode* pNode){if (pNode != NULL){cout << "value of this node is: " << pNode->m_nValue << endl;if (pNode->m_pLeft != NULL)cout << "value of its left child is: " << pNode->m_pLeft->m_nValue << endl;elsecout << "left child is null.\n";if (pNode->m_pRight != NULL)cout << "value of its right child is:" << pNode->m_pRight->m_nValue << endl;elsecout << "right child is null.\n";}else{cout << "this node is null.\n";}cout << endl;}//先序遍历二叉树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);}}//树深度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);}//逐个判断每个结点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);}//后序遍历的方法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;}