螺旋遍历

Problem statement:

问题陈述：

Write a program to print Level Order Traversal in spiral form of a binary tree.

编写一个程序以二叉树的螺旋形式打印Level Level Traversal 。

Example:

例：

For the above tree:
Basic level order traversal:
2
7 5
2 6 9
5 11 4
Level order traversal in spiral form:
2
7 5 (left to right)
9 6 2 (right to left)
5 11 4 (again left to right)

Solution:

解：

The solution will, of course, surround basic level order traversal. The spiral order means - It will go from left to right for one level, then right to left for next level and again left to right for the next one and so on.

当然，解决方案将围绕基本级别的顺序遍历 。 螺旋顺序表示-它将从左向右移动一个级别，然后从右向左移动到下一个级别，再从左向右移动到下一个级别，依此类推。

We need to modify our basic level order traversal.

我们需要修改基本的层级遍历 。

We can do the flipping of direction (left → right then right → left so on ...) by keeping a flag variable which will be updated at end of each level.

我们可以通过保持标记变量来进行方向翻转(左→右然后右→左等等)，这将在每个级别的末尾进行更新。

Pre-requisite: Root to tree

先决条件：从树到根

1.  Declare flag as 1(true);
2.  Declare a queue q to store pointer to nodes(node*);
3.  Declare a stack s which helps us for flipping.
4.  Print the root as we are not going to bother about root level;
5.  IF(root->left) //left child exists
ENQUEUE(q, root->left);
END IF
IF(root->right) //right child exists
ENQUEUE(q, root->right);
END IF
IF root has no child
RETURN BACK //nothing to print more
ELSE
q.push(NULL); //to indicate end of 1st level
6.  //Here goes the modified level order traversal
When flag=1 its left-to right 
flag=0 its right to left
while (q is not empty){
temp=DEQUEUE(q);	
IF(temp==NULL) //end of last traversed level
IF (q is not empty)
ENQUEUE (q, NULL);
END IF
IF (flag==0)
Pop and print data from stack until stack is empty
END IF	
flag=1-flag; //flip flag for next level 1 to 0 or 0 to 1
ELSE
IF(flag == 1)
Print temp->data; //left to right printing		
ELSE
Push temp->data to stack s; //this makes right to left 
//printing as rightmost node will be at the top of stack
END IF-ELSE
// basic level order traversal (direction left to right)
IF(root->left) //left child exists
ENQUEUE(q, root->left);
END IF
IF (root->right) //right child exists
ENQUEUE(q, root->right);
END IF
END IF-ELSE (outer one)
END WHILE loop

Example with Explanation:

解释示例：

    For the above tree root is being printed 
first without any constraint
2
For the first level flag is 1
Thus it prints immediately while accessing temp node
7 5 (since basic traversal direction is always left to right)
At the end of level flag flips to 0
So while traversing instead of printing nodes at once, 
nodes get stored in stack
At the end of level all the nodes data being popped and printed.
In stack
9(top)
6
2
Thus printing
9 6 2 (right to left)
Flag again flipped to 1
Basic left to right printing
5 11 4
So the output is in spiral order

C++ implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
// tree node is defined
class Node{
public:
int data;
Node *left;
Node *right;
};
// creating new node
Node* newnode(int data)  
{ 
Node* node = (Node*)malloc(sizeof(Node)); 
node->data = data; 
node->left = NULL; 
node->right = NULL; 
return(node); 
} 
void printSpiral(Node *root)
{
Node* temp;
int flag=1;	
queue<Node*> q;
stack<int> s;
cout<<root->data<<"\n";
if(root->left)
q.push(root->left);
if(root->right)
q.push(root->right);
if(!root->left && !root->right)
return;
q.push(NULL);
while(!q.empty()){
temp=q.front();
q.pop();
if(temp==NULL){
if(!q.empty())
q.push(NULL);
if(flag==0){
while(!s.empty()){
cout<<s.top()<<" ";
s.pop();
}
}
flag=1-flag;
cout<<endl;
}
else{
if(flag){
cout<<temp->data<<" ";
}
else{
s.push(temp->data);
}
if(temp->left)
q.push(temp->left);
if(temp->right)
q.push(temp->right);
}
}
}
int main() { 
//**same tree is builted as shown in example**
Node *root=newnode(2); 
root->left= newnode(7); 
root->right= newnode(5); 
root->right->right=newnode(9);
root->right->right->left=newnode(4);
root->left->left=newnode(2); 
root->left->right=newnode(6);
root->left->right->left=newnode(5);
root->left->right->right=newnode(11);
cout<<"Level Order traversal in spiral form of ";
cout<<"the binary tree is :"<<endl; 
printSpiral(root); 
return 0; 
}

Output

输出量

Level Order traversal in spiral form of the binary tree is :
2
7 5
9 6 2
5 11 4

翻译自: https://www.includehelp.com/icp/level-order-traversal-in-spiral-form.aspx

螺旋遍历