Binary Tree Node Structure in DSA Typescript

Mental Model

A binary tree node holds a value and links to two child nodes, one on the left and one on the right.

Analogy: Think of a family tree where each person can have up to two children, one on the left side and one on the right side.

    [Node]
    /    \
 [Left] [Right]
   null   null

Dry Run Walkthrough

Input: Create a binary tree node with value 5, left child 3, and right child 7

Goal: Build a node that connects to two child nodes correctly

Step 1: Create root node with value 5

[5] -> left: null, right: null

Why: Start with the main node that will hold the value

Step 2: Create left child node with value 3

[5] -> left: [3], right: null

Why: Add left child to hold smaller value

Step 3: Create right child node with value 7

[5] -> left: [3], right: [7]

Why: Add right child to hold larger value

Result:

[5] -> left: [3], right: [7]

Annotated Code

DSA Typescript

class TreeNode {
  value: number;
  left: TreeNode | null;
  right: TreeNode | null;

  constructor(value: number) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

// Create root node
const root = new TreeNode(5);
// Create left child
root.left = new TreeNode(3);
// Create right child
root.right = new TreeNode(7);

// Function to print tree node and its children
function printNode(node: TreeNode | null): string {
  if (!node) return "null";
  const leftVal = node.left ? node.left.value.toString() : "null";
  const rightVal = node.right ? node.right.value.toString() : "null";
  return `[${node.value}] -> left: [${leftVal}], right: [${rightVal}]`;
}

console.log(printNode(root));

this.left = null; this.right = null;

Initialize left and right child pointers to null

root.left = new TreeNode(3);

Assign left child node with value 3

root.right = new TreeNode(7);

Assign right child node with value 7

Output

[5] -> left: [3], right: [7]

Complexity Analysis

Time: O(1) because creating a node and assigning children is a constant time operation

Space: O(1) for each node created as it stores fixed number of references and a value

vs Alternative: Compared to arrays, nodes use pointers to link children which allows dynamic tree growth without resizing

Edge Cases

Node with no children

Both left and right pointers remain null indicating leaf node

DSA Typescript

this.left = null;
this.right = null;

Node with only left child

Right pointer stays null, left points to child node

DSA Typescript

root.left = new TreeNode(3);

Node with only right child

Left pointer stays null, right points to child node

DSA Typescript

root.right = new TreeNode(7);

When to Use This Pattern

When you need to represent hierarchical data with up to two children per item, use the binary tree node structure to link nodes simply and clearly.

Common Mistakes

Mistake: Forgetting to initialize left and right pointers to null
Fix: Always set left and right to null in the constructor to avoid undefined references

Mistake: Assigning children without creating new nodes
Fix: Create new TreeNode instances before assigning to left or right pointers

Summary

It creates a node that holds a value and links to two child nodes.

Use it when building trees where each node can have up to two children.

Remember each node has two pointers that start as null and can link to child nodes.