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.

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