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#include <iostream> #include <stack> using namespace std; #define MAX 100 //字符串最大长度 typedef struct Node //二叉树结点 { char data; Node *lchild,*rchild; } *Btree; void createBT(Btree &t); //先序构造二叉树 void preorder(Btree &t); //二叉树递归先序遍历 void inorder(Btree &t); //递归中序遍历 void postorder(Btree &t); //递归后序遍历 void preorderbyStack(Btree &t); //先序非递归遍历 void inorderbyStack(Btree &t); //中序非递归遍历 void postorderbyStack(Btree &t);//后序非递归遍历 /*****************最近公共父结点算法LCF*******************/ int findNode1(Btree &t,char ch,char p[]); //找出结点1到根的路径 int findNode2(Btree &t,char ch,char q[]); //找出结点2到根的路径 void compare(char *p,char *q); //比较两条路径,找到第一个公共结点 static int i = 0; static int j = 0; /*****************最近公共父结点算法LCF*******************/ void createBT(Btree &t) //先序创建二叉树 #表示空指针 { char ch; cin>>ch; if(ch == ‘#‘) t = NULL; else { if(!(t = new Node)) exit(1); t ->data = ch; //递归创建二叉树 createBT(t->lchild); createBT(t->rchild); } } void preorder(Btree &t) { if(t == NULL) return ; else { cout<<t->data<<" "; preorder(t->lchild); preorder(t->rchild); } } void inorder(Btree &t) { if(!t) return; else { inorder(t->lchild); cout<<t->data<<" "; inorder(t->rchild); } } void postorder(Btree &t) { if(!t) return; else { postorder(t->lchild); postorder(t->rchild); cout<<t->data<<" "; } } void preorderbyStack(Btree &t) { stack<Node*>s; Node *p = t; while (p || !s.empty()) { if(p) { cout<<p->data<<" "; //先序遍历二叉树,先访问根节点,再访问左子树 s.push(p); p = p->lchild; } else { p = s.top(); s.pop(); p = p->rchild; } } } void inorderbyStack(Btree &t) { stack<Node*>s; Node *p = t; while (p || !s.empty()) { if(p) { s.push(p); p = p->lchild; } else { p = s.top(); cout<<p->data<<" "; //中序遍历二叉树,先访问左子树 s.pop(); p = p->rchild; } } } void postorderbyStack(Btree &t) { stack<Node*> s; Node *p = t; Node *flag = nullptr; while (p || !s.empty()) { if(p) //后序遍历先访问左结点,再右结点,最后根结点 { s.push(p); p = p->lchild; } else { p = s.top(); if( p->rchild&&p->rchild != flag) //如果右子树不为空且没有被访问过 { p = p->rchild; //转到右子树 s.push(p); p = p->lchild; } else //如果右子树为空,当前结点为访问结点 { s.pop(); cout<<p->data<<" "; flag = p; //把刚刚访问的结点做标记,防止重复访问 p = nullptr; //把p置为空指针,如果不置位空。还要压入栈一次,就造成死循环 } } } } bool isEmpty(Btree &t) { return t == NULL?true:false; } int findNode1(Btree &t,char ch,char p[]) { if(!t) return 0; else if(t->data == ch) //从所给的结点出发,搜索到达根节点的路径,并将路径记录下来 { *(p+i) = t->data; ++i; return 1; } else { if(findNode1(t->lchild, ch, p)) { *(p+i) = t->data; ++i; return 1; } else if(findNode1(t->rchild, ch, p)) { *(p+i) = t->data; ++i; return 1; } else return 0; } } int findNode2(Btree &t,char ch,char q[]) { if(!t) return 0; else if(t->data == ch) { *(q+j) = t->data; ++j; return 1; } else { if(findNode2(t->lchild, ch, q)) { *(q+j) = t->data; ++j; return 1; } else if(findNode2(t->rchild, ch, q)) { *(q+j) = t->data; ++j; return 1; } else return 0; } } void compare(char *p,char *q) //比较两条路径,第一个不同的路径点的父结点就是其公共父结点 { i--; j--; while(p[i] == q[j]) { i--; j--; } cout<<"LCF is :"<<p[i+1]<<endl; } int main() { //测试用例:abc##de#g##f### Btree t = new Node; createBT(t); preorder(t); cout<<endl; preorderbyStack(t); cout<<endl; inorder(t); cout<<endl; inorderbyStack(t); cout<<endl; postorder(t); cout<<endl; postorderbyStack(t); cout<<endl; char a[MAX]; char b[MAX]; findNode1(t, ‘c‘,a); findNode2(t, ‘e‘,b); compare(a, b); return 0; }
前序构造二叉树 字符#表示空。
二叉树的递归遍历与非递归遍历。后者借助栈
最近公共父节点是指对于二叉树中的两个结点 找到最近的公共父节点
二叉树递归与非递归遍历,最近公共父节点算法,布布扣,bubuko.com
标签:blog os io ar div amp log new
原文地址:http://www.cnblogs.com/cliviazhou/p/3916924.html