python中怎么分析循环遍历二叉树，很多新手对此不是很清楚，为了帮助大家解决这个难题，下面小编将为大家详细讲解，有这方面需求的人可以来学习下，希望你能有所收获。
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前序遍历
struct Node
{
Node*left;
Node*right;
int data;
Node(){ func; }
};
Node* create(Node*p, int depth)
{ if (p && depth)
{
p->left = new Node;
p->right = new Node;
p->data = depth;
create(p->left, depth - 1);
create(p->right, depth - 1);
} if (!depth)
{
p->left = nullptr;
p->right = nullptr;
p->data = depth;
} return p;
}
void print1(Node*p)
{ if (p)
{
cout << p->data << " ";
print1(p->left);
print1(p->right);
}
}
void print2(Node*head)//利用stack 模拟函数调用过程 来遍历{
stack s;
Node*p = head;
{ while (p)
{
s.push(p);
cout << p->data << " ";
p = p->left;
} while (!s.empty())
{
Node*pp = s.top(); if (pp->right && pp != head)
{
cout << pp->right->data << " ";
}
s.pop();
}
}
{
p = head->right; while (p)
{
s.push(p);
cout << p->data << " ";
p = p->left;
} while (!s.empty())
{
Node*pp = s.top(); if (pp->right
&& pp != head)
{
cout << pp->right->data << " ";
}
s.pop();
}
}
}
int main()
{
Node* head = new Node;
create(head, 2);
head->data = 10;
head->left->data = 6;
head->right->data = 14;
head->left->left->data = 4;
head->left->right->data = 8;
head->right->left->data = 12;
head->right->right->data = 16;
print1(head);//递归遍历
cout << endl;
print2(head);//循环遍历
system("pause"); return 0;
} 中序
void print2(Node*head)
{
stack s;
Node*p = head;
{ while (p)
{
s.push(p); // cout << p->data << " ";
p = p->left;
} while (!s.empty())
{
Node*pp = s.top();
cout << pp->data << " "; if (pp->right && pp != head )
{
cout << pp->right->data << " ";
}
s.pop();
}
}
{
p = head->right; while (p)
{
s.push(p);
p = p->left;
} while (!s.empty())
{
Node*pp = s.top();
cout << pp->data << " "; if (pp->right&& pp != head)
{
cout << pp->right->data << " ";
}
s.pop();
}
}
} 后序
void print2(Node*head)
{
stack s;
Node*p = head;
{ while (p)
{
s.push(p);
p = p->left;
} while (!s.empty())
{
Node*pp = s.top();
if (pp->right && pp != head )
{
cout << pp->right->data << " ";
}
if ( pp != head)
cout << pp->data << " ";
s.pop();
}
}
{
p = head->right; while (p)
{
s.push(p);
p = p->left;
} while (!s.empty())
{
Node*pp = s.top();
if (pp->right&& pp != head)
{
cout << pp->right->data << " ";
}
cout << pp->data << " ";
s.pop();
}
}
cout << head->data << " ";
} 看完上述内容是否对您有帮助呢？如果还想对相关知识有进一步的了解或阅读更多相关文章，请关注创新互联行业资讯频道，感谢您对创新互联的支持。
