Question
Consider the following Java-like pseudo-code for
inserting a node into a Binary Search Tree (BST):   ```java   class Node {     int data;     Node left, right;     public Node(int item) {       data = item;       left = right = null;     }   }   class BST {     Node root;     BST() {       root = null;     }     void insert(int data) {       root = insertRec(root, data);     }     Node insertRec(Node root, int data) {       if (root == null) {         root = new Node(data);         return root;       }       if (data < root.data) {         root.left = insertRec(root.left, data);       } else if (data > root.data) {         root.right = insertRec(root.right, data);       }       // If data == root.data, do nothing (assume no duplicates)       return root;     }   }   ```   If you insert the following sequence of numbers into an initially empty BST: `50, 30, 70, 20, 40, 60, 80`, what will be the data of the node that has `40` as its right child?ÂSolution
Let's trace the BST construction:     1. Insert `50`: `root = 50`     2. Insert `30`: `30 < 50`, so `30` becomes `50`'s left child.       ```         50         /        30       ```     3. Insert `70`: `70 > 50`, so `70` becomes `50`'s right child.       ```         50         / \        30  70       ```     4. Insert `20`: `20 < 50`, go left to `30`. `20 < 30`, so `20` becomes `30`'s left child.       ```         50         / \        30  70        /       20       ```     5. Insert `40`: `40 < 50`, go left to `30`. `40 > 30`, so `40` becomes `30`'s right child.       ```         50         / \        30  70        / \       20 40       ```     6. Insert `60`: `60 > 50`, go right to `70`. `60 < 70`, so `60` becomes `70`'s left child.       ```         50         / \        30  70        / \ /       20 40 60       ```     7. Insert `80`: `80 > 50`, go right to `70`. `80 > 70`, so `80` becomes `70`'s right child.       ```         50         / \        30  70        / \ / \       20 40 60 80       ```     The node that has `40` as its right child is `30`.
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