{"post":"

Level order traversal in Spiral form

We will traverse each level and store all the nodes at that level in a temporary vector. We will keep the current direction in a variable that will change after traversing each level.

#include<bits/stdc++.h>\nusing namespace std;\nstruct Node{\nint data;\nNode* left;\nNode* right;\nNode(int data){\nthis->data = data;\nleft = right = NULL;\n}\n};
\nvoid levelOrderSpiral(Node* root, int state){\nif(!root) return;
\nqueue&lt;Node*&gt;q;\nq.push(root);\nbool dir = 0;\nwhile(!q.empty()){\n\tint size = q.size();\n  \tvector&lt;Node*&gt; temp;\n  \tfor(int i = 0; i &lt; size; i++){\n    \tNode* curr = q.front();\n      \tq.pop();\n      \ttemp.push_back(curr);\n      \tif(curr-&gt;left) q.push(curr-&gt;left);\n        if(curr-&gt;right) q.push(curr-&gt;right);\n    }\n  \tif(dir == 0){\n    \tfor(auto i:temp) cout&lt;&lt;i-&gt;data&lt;&lt;\" \";\n    } else {\n    \tfor(int i = temp.size()-1; i &gt;= 0; i--){\n        \tcout&lt;&lt;temp[i]-&gt;data&lt;&lt;\" \";\n        }\n    }\n  \tdir = !dir;\n}\n
\n}
\nint main() {\n/*\n1\n/    
\n2      3\n/   \\   /  
\n4     5 6    7\n/\nstruct Node root = new Node(1);\nroot->left = new Node(2);\nroot->left->left = new Node(4);\nroot->left->right = new Node(5);\nroot->right = new Node(3);\nroot->right->left = new Node(6);\nroot->right->right = new Node(7);\nlevelOrderSpiral(root, 0);\nreturn 0;\n}

\n","title":"Level order traversal – Spiral form","authorDetails":{"name":"studymite","slug":"admin"},"description":"Here we will do a level order traversal of a tree in spiral form using breadth first search technique.","slug":"level-order-traversal-spiral-form/","articleId":"636a83417bb1ce5072622b02","bookmarkStatus":{"isBookmarked":false,"bookmarkId":null}}