求二叉树的镜像

6-02 1,936 views

思路：
采用递归的思想。对于根节点，若其左子树或右子树不为空，则互换左、右子树，然后对于左、右子树，分别递归上述处理方法，直至叶节点。
示例：

代码：

typedef struct TNode
{
    int value;
    TNode* lchild;
    TNode* rchild;
}TNode,*BTree; 

//采用递归进行镜像转换
void MirrorTree(BTree tree)
{
    if (tree == NULL) return;
    if (tree->lchild == NULL && tree->rchild == NULL) return;
    TNode* temp = tree->lchild;
    tree->lchild = tree->rchild;
    tree->rchild = temp;
    MirrorTree(tree->lchild);
    MirrorTree(tree->rchild);
}

typedef struct TNode

{

int value;

TNode* lchild;

TNode* rchild;

}TNode,*BTree;

//采用递归进行镜像转换

void MirrorTree(BTree tree)

{

if (tree == NULL) return;

if (tree->lchild == NULL && tree->rchild == NULL) return;

TNode* temp = tree->lchild;

tree->lchild = tree->rchild;

tree->rchild = temp;

MirrorTree(tree->lchild);

MirrorTree(tree->rchild);

}

版权属于: 我爱我家

原文地址: http://magicwt.com/2012/06/02/%e6%b1%82%e4%ba%8c%e5%8f%89%e6%a0%91%e7%9a%84%e9%95%9c%e5%83%8f/

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