node to queue as it represents the first level queue. For example, in the following tree, sum of nodes at odd level is ( ) which. Left new Node(2. If(currentLevel 2 0) /If level is even, add nodes's to variable evenLevel evenLevel current- data; else /If level is odd, add nodes's to variable oddLevel oddLevel current- data; /Adds left child to queue if(current- left! Right new Node(9. EvenLevelSum 2 3 5, difference, algorithm. Data else /If level is odd, add nodes's to variable oddLevel oddLevel Int32(current.
Consider root as level 1, left and right.
Find depth of the deepest odd level leaf node.
Print Postorder traversal from given Inorder and Preorder traversals.
You can do this easier if you break the problem in 2 different functions (not very efficient) / Method to calculate the even or odd depending on the flag public static int even_odd (Node n, int level, int flag) int value 0; if (n null). Recommended Posts: Article Tags : Tree Amazon thumb_up 1 To-do Done.4 Based on 93 vote(s) Please write to us at to report any issue with the above content. And sum of nodes at even level is ( ) which. 0) /Variable nodesInLevel will hold the size of queue.e. Calculate the difference by subtracting value present in evenLevel from oddLevel. 0 #Variable nodesInLevel will hold the size of queue.e. Data /Adds left child to queue if(current. Please use ideone or, c Shell or any other online compiler link to post code in comments. Please spread the word and help us grow.
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