編程演算法 - 判斷二叉樹是不是平衡樹 代碼(C)

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判斷二叉樹是不平衡樹 代碼(C)

本文地址: http://blog.csdn.net/caroline_wendy

題目: 輸入一顆二叉樹的根結點, 判斷該樹是不是平衡二叉樹.

二叉平衡樹: 任意結點的左右子樹的深度相差不超過1.

使用後序遍曆的方式, 並且儲存左右子樹的深度, 進行比較.

代碼:

/* * main.cpp * *  Created on: 2014.6.12 *      Author: Spike *//*eclipse cdt, gcc 4.8.1*/#include <stdio.h>#include <stdlib.h>#include <string.h>struct BinaryTreeNode {int m_nValue;BinaryTreeNode* m_pLeft;BinaryTreeNode* m_pRight;};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(BinaryTreeNode* pRoot) {int depth = 0;return IsBalanced(pRoot, &depth);}BinaryTreeNode* init(void) {BinaryTreeNode* pRoot = new BinaryTreeNode(); pRoot->m_nValue = 1;BinaryTreeNode* pNode2 = new BinaryTreeNode(); pNode2->m_nValue = 2;BinaryTreeNode* pNode3 = new BinaryTreeNode(); pNode3->m_nValue = 3;BinaryTreeNode* pNode4 = new BinaryTreeNode(); pNode4->m_nValue = 4;BinaryTreeNode* pNode5 = new BinaryTreeNode(); pNode5->m_nValue = 5;BinaryTreeNode* pNode6 = new BinaryTreeNode(); pNode6->m_nValue = 6;BinaryTreeNode* pNode7 = new BinaryTreeNode(); pNode7->m_nValue = 7;pRoot->m_pLeft = pNode2; pRoot->m_pRight = pNode3;pNode2->m_pLeft = pNode4; pNode2->m_pRight = pNode5;pNode4->m_pLeft = NULL; pNode4->m_pRight = NULL;pNode5->m_pLeft = pNode7; pNode5->m_pRight = NULL;pNode7->m_pLeft = NULL; pNode7->m_pRight = NULL;pNode3->m_pLeft = NULL; pNode3->m_pRight = pNode6;pNode6->m_pLeft = NULL; pNode6->m_pRight = NULL;return pRoot;}int main(void){BinaryTreeNode* pRoot = init();bool result = IsBalanced(pRoot);printf("result = %s\n", result==false?"false":"true");    return 0;}

輸出:

result = true