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剑指offer第三十九题-拓展：输入一棵二叉树的根结点，判断该树是不是平衡二叉树（AVL）
1 //============================================================================ 2 // Name : JZ-C-39-Plus.cpp 3 // Author : Laughing_Lz 4 // Version : 5 // Copyright : All Right Reserved 6 // Description : 39-拓展：输入一棵二叉树的根结点，判断该树是不是平衡二叉树（AVL） 7 //============================================================================ 8 9 #include <iostream> 10 #include <stdio.h> 11 #include "BinaryTree.h" 12 using namespace std; 13 14 // ====================方法1==================== 15 /** 16 * 代码简洁，但是由于一个结点会被重复遍历多次，所以效率不高 17 */ 18 int TreeDepth(BinaryTreeNode* pRoot) { 19 if (pRoot == NULL) 20 return 0; 21 22 int nLeft = TreeDepth(pRoot->m_pLeft); 23 int nRight = TreeDepth(pRoot->m_pRight); 24 25 return (nLeft > nRight) ? (nLeft + 1) : (nRight + 1); 26 } 27 28 bool IsBalanced_Solution1(BinaryTreeNode* pRoot) { 29 if (pRoot == NULL) 30 return true; 31 32 int left = TreeDepth(pRoot->m_pLeft); 33 int right = TreeDepth(pRoot->m_pRight); 34 int diff = left - right; 35 if (diff > 1 || diff < -1) 36 return false; 37 38 return IsBalanced_Solution1(pRoot->m_pLeft) 39 && IsBalanced_Solution1(pRoot->m_pRight); 40 } 41 42 // ====================方法2==================== 43 /** 44 * 采用后序遍历★方式遍历二叉树的每一个结点，在遍历到一个结点之前就已经遍历了它的左右子树。 45 * 一边遍历一边判断每个结点是不是平衡的 46 */ 47 bool IsBalanced(BinaryTreeNode* pRoot, int* pDepth); 48 49 bool IsBalanced_Solution2(BinaryTreeNode* pRoot) { 50 int depth = 0; 51 return IsBalanced(pRoot, &depth); 52 } 53 54 bool IsBalanced(BinaryTreeNode* pRoot, int* pDepth) { 55 if (pRoot == NULL) { 56 *pDepth = 0; 57 return true; 58 } 59 60 int left, right; 61 if (IsBalanced(pRoot->m_pLeft, &left) 62 && IsBalanced(pRoot->m_pRight, &right)) { 63 int diff = left - right; 64 if (diff <= 1 && diff >= -1) { 65 *pDepth = 1 + (left > right ? left : right); 66 return true; 67 } 68 } 69 70 return false; 71 } 72 73 // ====================测试代码==================== 74 void Test(char* testName, BinaryTreeNode* pRoot, bool expected) { 75 if (testName != NULL) 76 printf("%s begins:\n", testName); 77 78 printf("Solution1 begins: "); 79 if (IsBalanced_Solution1(pRoot) == expected) 80 printf("Passed.\n"); 81 else 82 printf("Failed.\n"); 83 84 printf("Solution2 begins: "); 85 if (IsBalanced_Solution2(pRoot) == expected) 86 printf("Passed.\n"); 87 else 88 printf("Failed.\n"); 89 } 90 91 // 完全二叉树 92 // 1 93 // / 94 // 2 3 95 // /\ / 96 // 4 5 6 7 97 void Test1() { 98 BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1); 99 BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2); 100 BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3); 101 BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4); 102 BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5); 103 BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6); 104 BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7); 105 106 ConnectTreeNodes(pNode1, pNode2, pNode3); 107 ConnectTreeNodes(pNode2, pNode4, pNode5); 108 ConnectTreeNodes(pNode3, pNode6, pNode7); 109 110 Test("Test1", pNode1, true); 111 112 DestroyTree(pNode1); 113 } 114 115 // 不是完全二叉树，但是平衡二叉树 116 // 1 117 // / 118 // 2 3 119 // /\ 120 // 4 5 6 121 // / 122 // 7 123 void Test2() { 124 BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1); 125 BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2); 126 BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3); 127 BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4); 128 BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5); 129 BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6); 130 BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7); 131 132 ConnectTreeNodes(pNode1, pNode2, pNode3); 133 ConnectTreeNodes(pNode2, pNode4, pNode5); 134 ConnectTreeNodes(pNode3, NULL, pNode6); 135 ConnectTreeNodes(pNode5, pNode7, NULL); 136 137 Test("Test2", pNode1, true); 138 139 DestroyTree(pNode1); 140 } 141 142 // 不是平衡二叉树 143 // 1 144 // / 145 // 2 3 146 // /147 // 4 5 148 // / 149 // 6 150 void Test3() { 151 BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1); 152 BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2); 153 BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3); 154 BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4); 155 BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5); 156 BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6); 157 158 ConnectTreeNodes(pNode1, pNode2, pNode3); 159 ConnectTreeNodes(pNode2, pNode4, pNode5); 160 ConnectTreeNodes(pNode5, pNode6, NULL); 161 162 Test("Test3", pNode1, false); 163 164 DestroyTree(pNode1); 165 } 166 167 // 1 168 // / 169 // 2 170 // / 171 // 3 172 // / 173 // 4 174 // / 175 // 5 176 void Test4() { 177 BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1); 178 BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2); 179 BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3); 180 BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4); 181 BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5); 182 183 ConnectTreeNodes(pNode1, pNode2, NULL); 184 ConnectTreeNodes(pNode2, pNode3, NULL); 185 ConnectTreeNodes(pNode3, pNode4, NULL); 186 ConnectTreeNodes(pNode4, pNode5, NULL); 187 188 Test("Test4", pNode1, false); 189 190 DestroyTree(pNode1); 191 } 192 193 // 1 194 // 195 // 2 196 // 197 // 3 198 // 199 // 4 200 // 201 // 5 202 void Test5() { 203 BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1); 204 BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2); 205 BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3); 206 BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4); 207 BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5); 208 209 ConnectTreeNodes(pNode1, NULL, pNode2); 210 ConnectTreeNodes(pNode2, NULL, pNode3); 211 ConnectTreeNodes(pNode3, NULL, pNode4); 212 ConnectTreeNodes(pNode4, NULL, pNode5); 213 214 Test("Test5", pNode1, false); 215 216 DestroyTree(pNode1); 217 } 218 219 // 树中只有1个结点 220 void Test6() { 221 BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1); 222 Test("Test6", pNode1, true); 223 224 DestroyTree(pNode1); 225 } 226 227 // 树中没有结点 228 void Test7() { 229 Test("Test7", NULL, true); 230 } 231 232 int main(int argc, char** argv) { 233 Test1(); 234 Test2(); 235 Test3(); 236 Test4(); 237 Test5(); 238 Test6(); 239 Test7(); 240 241 return 0; 242 }
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原文地址：http://www.cnblogs.com/Laughing-Lz/p/5608553.html
