标签:mystra 编程算法 判断二叉树是不是平衡树 代码 c
本文地址: http://blog.csdn.net/caroline_wendy
题目: 输入一颗二叉树的根结点, 判断该树是不是平衡二叉树.
二叉平衡树: 任意结点的左右子树的深度相差不超过1.
使用后序遍历的方式, 并且保存左右子树的深度, 进行比较.
代码:
/* * 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; }
result = true
编程算法 - 判断二叉树是不是平衡树 代码(C),布布扣,bubuko.com
标签:mystra 编程算法 判断二叉树是不是平衡树 代码 c
原文地址:http://blog.csdn.net/caroline_wendy/article/details/36873523