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DSA Typescriptprogramming~5 mins

Tree Traversal Preorder Root Left Right in DSA Typescript - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is preorder tree traversal?
Preorder traversal visits nodes in this order: Root first, then Left subtree, and finally Right subtree. It means you process the current node before its children.
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beginner
In preorder traversal, which node is visited first?
The root node is visited first in preorder traversal before visiting any child nodes.
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beginner
What is the order of visiting nodes in preorder traversal?
The order is: Root → Left subtree → Right subtree.
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intermediate
Why is preorder traversal useful?
Preorder traversal is useful to copy a tree or to get prefix expression of an expression tree because it processes the root before its children.
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beginner
Show the preorder traversal output for this tree:

1
/ \ 2 3
/ / \ 4 5 6
Preorder traversal visits nodes as: 1 → 2 → 4 → 3 → 5 → 6
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In preorder traversal, which node is visited immediately after the root?
AParent node
BRight child
CLeft child
DLeaf node
What is the correct order of nodes visited in preorder traversal?
ALeft → Root → Right
BRight → Left → Root
CLeft → Right → Root
DRoot → Left → Right
Which traversal visits the root node before its children?
AInorder
BPreorder
CPostorder
DLevel order
If a tree has only one node, what is the preorder traversal output?
AThe single node
BEmpty
CNull
DDepends on children
Which of these is NOT a use case of preorder traversal?
ASorting nodes in ascending order
BPrefix expression of expression tree
CCopying a tree
DVisiting root before children
Explain how preorder traversal works and the order in which nodes are visited.
Think about visiting the root before going down to children.
You got /4 concepts.
    Describe a real-life example where preorder traversal would be useful.
    Consider when you need to process a parent before its children.
    You got /4 concepts.

      Practice

      (1/5)
      1. What is the order of nodes visited in a preorder tree traversal?
      easy
      A. Root, Left, Right
      B. Left, Root, Right
      C. Left, Right, Root
      D. Right, Left, Root

      Solution

      1. Step 1: Understand preorder traversal definition

        Preorder traversal visits the root node first, then recursively visits the left subtree, followed by the right subtree.
      2. Step 2: Match the order with options

        The order Root, Left, Right matches the preorder traversal definition exactly.
      3. Final Answer:

        Root, Left, Right -> Option A
      4. Quick Check:

        Preorder = Root, Left, Right [OK]
      Hint: Preorder always starts with the root node first [OK]
      Common Mistakes:
      • Confusing preorder with inorder or postorder
      • Visiting left or right before root
      • Mixing up traversal orders
      2. Which of the following TypeScript function signatures correctly represents a preorder traversal function for a binary tree node?
      easy
      A. function preorder(node: TreeNode): number[]
      B. function preorder(node: TreeNode | null): void
      C. function preorder(): TreeNode
      D. function preorder(node: TreeNode, result: number[]): number

      Solution

      1. Step 1: Analyze function parameters and return type

        A preorder traversal function typically takes a node (or null) and returns void if it prints or processes nodes directly.
      2. Step 2: Check each option

        function preorder(node: TreeNode | null): void accepts a node or null and returns void, which fits a traversal that processes nodes without returning a value.
      3. Final Answer:

        function preorder(node: TreeNode | null): void -> Option B
      4. Quick Check:

        Traversal function signature with node or null and void return [OK]
      Hint: Traversal functions usually take a node and return void or array [OK]
      Common Mistakes:
      • Missing null in node parameter type
      • Wrong return type for traversal
      • Incorrect number of parameters
      3. Given the following binary tree:
          1
         / \
        2   3
       / \
      4   5

      What is the output of a preorder traversal (Root, Left, Right)?
      medium
      A. [4, 2, 5, 1, 3]
      B. [2, 4, 5, 1, 3]
      C. [1, 2, 4, 5, 3]
      D. [1, 3, 2, 4, 5]

      Solution

      1. Step 1: Start at root node 1

        Preorder visits root first: 1
      2. Step 2: Traverse left subtree of 1

        Visit node 2, then its left child 4, then right child 5: 2, 4, 5
      3. Step 3: Traverse right subtree of 1

        Visit node 3: 3
      4. Final Answer:

        [1, 2, 4, 5, 3] -> Option C
      5. Quick Check:

        Preorder = Root, Left, Right = [1, 2, 4, 5, 3] [OK]
      Hint: Visit root first, then left subtree, then right subtree [OK]
      Common Mistakes:
      • Visiting left subtree after right subtree
      • Starting traversal from left child instead of root
      • Mixing preorder with inorder or postorder
      4. Consider this TypeScript preorder traversal function:
      function preorder(node: TreeNode | null): void {
        if (node === null) return;
        preorder(node.left);
        console.log(node.val);
        preorder(node.right);
      }

      What is the error in this code?
      medium
      A. It does not handle null nodes
      B. It will cause infinite recursion
      C. It prints nodes in postorder
      D. It visits nodes in inorder, not preorder

      Solution

      1. Step 1: Analyze the order of operations in the function

        The function first calls preorder on left child, then prints current node, then calls preorder on right child.
      2. Step 2: Compare with preorder definition

        Preorder should print the node first, then traverse left and right. This code prints after left subtree, so it is inorder traversal.
      3. Final Answer:

        It visits nodes in inorder, not preorder -> Option D
      4. Quick Check:

        Print before recursive calls = preorder; here print is in middle [OK]
      Hint: Print node value before recursive calls for preorder [OK]
      Common Mistakes:
      • Printing node after left subtree instead of before
      • Confusing traversal orders
      • Missing base case for null
      5. You want to collect all node values of a binary tree in preorder traversal into an array. Which TypeScript function correctly implements this?
      hard
      A. function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; }
      B. function preorder(node: TreeNode | null): number[] { if (!node) return []; return [...preorder(node.left), node.val, ...preorder(node.right)]; }
      C. function preorder(node: TreeNode | null): number[] { if (!node) return []; return [...preorder(node.left), ...preorder(node.right), node.val]; }
      D. function preorder(node: TreeNode | null): number[] { if (!node) return []; return [node.val, ...preorder(node.right), ...preorder(node.left)]; }

      Solution

      1. Step 1: Understand preorder traversal collection

        Preorder visits root first, then left subtree, then right subtree, collecting values in order.
      2. Step 2: Check each function's order and correctness

        function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; } pushes node.val first, then recursively collects left and right, returning the result array correctly.
      3. Final Answer:

        function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; } -> Option A
      4. Quick Check:

        Push root first, then left and right recursively [OK]
      Hint: Push root value before recursive calls to left and right [OK]
      Common Mistakes:
      • Collecting left subtree before root
      • Mixing left and right subtree order
      • Returning new arrays instead of accumulating
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