中序遍历二叉树
中序遍历二叉树的非递归实现:
先不停往左孩子方向走,一边往左走一边入栈,等走到尽头后输出一个结点来访问,然后走向这个结点的右孩子。在右孩子结点非空或栈非空的状态下,不停迭代上述过程。
依旧是这棵老树:
/*------------
树的形状:
a
b c
d e f
--------------*/
代码如下:
#include<iostream>
#include<stack>
#include<queue>
using namespace std;
typedef struct node{
char val;
struct node* left;
struct node* right;
}TreeNode,*Tree;
//中序遍历(递归)
void InOrder(Tree& t){
if(t == NULL) return;
//递归遍历左子树
InOrder(t->left);
cout<<t->val<<" ";//输出当前结点的值
//递归遍历右子树
InOrder(t->right);
}
//中序遍历(非递归)
void InOrderTraversal(Tree& t){
stack<TreeNode*> s; //创建一个树结点型的栈
TreeNode* p = t;
while(p != NULL || !s.empty()){
//左孩子不为空,一直往左走
while(p != NULL){
s.push(p);
p=p->left;
}
//出栈访问
p = s.top();
cout<<p->val<<" ";
s.pop();
//走向右孩子
p = p->right;
}
}
void CreateTree(Tree& t){
char x;
cin>>x;
if(x == '#') t = NULL;
else{
t = new TreeNode;
t->val = x;
CreateTree(t->left);
CreateTree(t->right);
}
}
void levelOrder(Tree& t) {
if(t == NULL) return;
queue<TreeNode*> q;
q.push(t);
while(!q.empty()){
int n = q.size();
for(int i = 0;i<n;i++){
TreeNode* s = q.front();
cout<<s->val<<" ";
q.pop();
if(s->left) q.push(s->left);
if(s->right) q.push(s->right);
}
cout<<endl;
}
}
int main(){
Tree t;
CreateTree(t);
/*
a b d # # e # # c f # # #
*/
levelOrder(t);
cout<<endl<<"递归:"<<endl;
InOrder(t);
cout<<endl<<"非递归:"<<endl;
InOrderTraversal(t);
}
运行结果: