【剑指offer】判断二叉树平衡

来源：互联网发布：幸福哲学书知乎编辑：程序博客网时间：2024/05/16 06:14

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题目：输入一个二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节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。

剑指offer上给的第二种思路，用后序遍历真的是将递归发挥的淋漓尽致，先遍历节点的左右子树，左右子树都平衡才来判断该节点是否平衡，如果左右子树中有不平衡的，则直接返回false，避免了从上往下逐个节点地计算深度带来的重复遍历节点。

代码如下：

#include<stdio.h>#include<stdlib.h>typedef struct BTNode{char data;struct BTNode *pLchild;struct BTNode *pRchild;}BTNode, *BTree;BTree create_tree();bool IsBalanced(BTree,int *);bool IsBalanced(BTree);int main(){BTree pTree = create_tree();if(IsBalanced(pTree))printf("Balanced\n");elseprintf("Not Balanced\n");return 0;}BTree create_tree(){BTree pA = (BTree)malloc(sizeof(BTNode));BTree pB = (BTree)malloc(sizeof(BTNode));BTree pD = (BTree)malloc(sizeof(BTNode));BTree pE = (BTree)malloc(sizeof(BTNode));BTree pC = (BTree)malloc(sizeof(BTNode));BTree pF = (BTree)malloc(sizeof(BTNode));pA->data = 'A';pB->data = 'B';pD->data = 'D';pE->data = 'E';pC->data = 'C';pF->data = 'F';pA->pLchild = pB;pA->pRchild = pC;pB->pLchild = pD;pB->pRchild = pE;pD->pLchild = NULL;pD->pRchild = NULL;pE->pLchild = pE->pRchild = NULL;pC->pLchild = NULL;pC->pRchild = pF;pF->pLchild = pF->pRchild = NULL;return pA;}/*后续递归遍历判断二叉树是否平衡*/bool IsBalanced(BTree pTree,int *depth){if(pTree == NULL){*depth = 0;return true;}int leftDepth,rightDepth;if(IsBalanced(pTree->pLchild,&leftDepth) && IsBalanced(pTree->pRchild,&rightDepth)){int diff = leftDepth-rightDepth;if(diff<=1 && diff>=-1){*depth = (leftDepth>rightDepth ? leftDepth:rightDepth) + 1;return true;}}return false;}/*封装起来*/bool IsBalanced(BTree pTree){int depth = 0;return IsBalanced(pTree,&depth);}

测试结果：

3 0