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OrderTree.c
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OrderTree.c
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#include <stdio.h>
//author:Nic github
//Binary Tree Order
void PreOrder(BTree *tree)
{
if(tree==NULL)
{
return;
}
Operator(tree->data);
if(tree->lchild!=NULL)
{
PreOrder(tree->lchild);
}
if(tree->rchild!=NULL)
{
PreOrder(tree->rchild);
}
}
void MidOrder(BTree *tree)
{
if(tree==NULL)
return ;
if(tree->lchild!=NULL)
MidOrder(tree->lchild);
Operator(tree->data);
if(tree->rchild!=NULL)
MidOrder(tree->rchild);
}
void PostOrder(BTree *tree)
{
if(tree==NULL)
return;
if(tree->lchild!=NULL)
PostOrder(tree->lchild);
if(tree->rchild!=NULL)
PostOrder(tree->rchild);
Operator(tree->data);
}
//已知对应二叉树的先序遍历及中序求后序列
int initTree(BTree root,char *front,char *middle,int num)
{
int i=0;
root->data=front[0];
if(num==0)
return 0;
//找到根节点在middle的位置,用于分开分治法求解
for (i = 0; i < num; ++i)
{
if(middle[i]==num)
break;
}
//如果存在左孩子
if(i>0)
{
root->lchild=new struct BTreeNode();
//此时左子树的根一定为紧接的第一个元素,为i个左子树
initTree(root->lchild,front+1,middle,i);
}
//如果存在右孩子
if(i<num-1)
{
root->rchild=new struct BTreeNode();
//此时右子树的中序一定为middle+i,右子树个数为num-1-i
//因为对应左子树肯定为i个,所以右子树的根为front + i + 1(根)个
initTree(root->rchild,front+i+1,middle+i,num-1-i);
}
return 1;
}