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  1. Prepare
  2. Data Structures
  3. Trees
  4. Tree: Inorder Traversal
  5. Discussions

Tree: Inorder Traversal

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212 Discussions

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  • bindra_sheetal05
     + 0 comments

    PYTHON3 Solution:

    def inOrder(root):
    #Write your code here
     if root == None:
         return
     inOrder(root.left)
     print(root.info, end=" ")
     inOrder(root.right)

  • emhhacker
     + 0 comments

    My Java solution with linear time complexity and o(h) space complexity:

    public static void inOrder(Node root) {
        if(root == null) return;
        
        inOrder(root.left);
        System.out.print(root.data + " ");
        inOrder(root.right);
    }

  • yahel_codes
     + 0 comments

    C 

    void inOrder( struct node *root) {
    if(!root) return;
    inOrder(root->left);
    printf("%d ", root->data); 
    inOrder(root->right);
}

  • calancea_bianca1
     + 0 comments

    Python

    def inOrder(root):
    if root is None:
        return
    inOrder(root.left)
    print(root.info,end=" ")
    inOrder(root.right)

  • jmcparland
     + 0 comments

    Haskell

    module Main where

data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show)

inOrder :: Tree a -> [a]
inOrder Empty = []
inOrder (Node a left right) = inOrder left ++ [a] ++ inOrder right

insertNode :: (Ord a) => Tree a -> a -> Tree a
insertNode Empty x = Node x Empty Empty
insertNode (Node a left right) x
    | x < a = Node a (insertNode left x) right
    | x > a = Node a left (insertNode right x)
    | otherwise = Node a left right

main :: IO ()
main =
    interact $
        unwords . map show . inOrder . foldl insertNode Empty . map (read :: String -> Int) . words . (!! 1) . lines