Merge pull request #8 from rvguradiya/full-binary-tree-conversion

convert a binary tree to full tree by removing node that have only on…

Author: rvguradiya

problems/tree/full-binary-tree-conversion/Approach.java

@@ -0,0 +1,65 @@
+class TreeNode {
+    int value;
+    TreeNode left, right;
+
+    public TreeNode(int value) {
+        this.value = value;
+        left = right = null;
+    }
+}
+
+public class Approach {
+
+    // Method to convert the binary tree to a full binary tree
+    public static TreeNode convertToFullBinaryTree(TreeNode node) {
+        TreeNode tempL = null, tempR = null, temp = null;
+        if (node.left == null && node.right == null) {
+            return node;
+        }
+        if (node.left != null) {
+            tempL = convertToFullBinaryTree(node.left);
+        }
+        if (node.right != null) {
+            tempR = convertToFullBinaryTree(node.right);
+        }
+        if (node.left != null && node.right != null) {
+            node.left = tempL;
+            node.right = tempR;
+            return node;
+        }
+        return tempL == null ? tempR : tempL;
+    }
+
+    // Method to print the tree in InOrder (used for testing)
+    public static void printInOrder(TreeNode node) {
+        if (node == null) {
+            return;
+        }
+        printInOrder(node.left);
+        System.out.print(node.value + " ");
+        printInOrder(node.right);
+    }
+
+    public static void main(String[] args) {
+        // Construct the binary tree
+        TreeNode root = new TreeNode(0);
+        root.left = new TreeNode(1);
+        root.right = new TreeNode(2);
+        root.left.left = new TreeNode(3);
+        root.left.left.right = new TreeNode(5);
+        root.right.right = new TreeNode(4);
+        root.right.right.left = new TreeNode(6);
+        root.right.right.right = new TreeNode(7);
+
+        System.out.println("Original Binary Tree (InOrder):");
+        printInOrder(root);
+        System.out.println();
+
+        // Convert to full binary tree
+        TreeNode fullTree = convertToFullBinaryTree(root);
+
+        System.out.println("Full Binary Tree (InOrder):");
+        printInOrder(fullTree);
+        System.out.println();
+    }
+}

problems/tree/full-binary-tree-conversion/README.md

@@ -0,0 +1,26 @@
+This problem was asked by **Yahoo**.
+
+Recall that a full binary tree is one in which each node is either a leaf node, or has two children. Given a binary tree, convert it to a full one by removing nodes with only one child.
+
+For example, given the following tree:
+
+```graph
+
+         0
+      /     \
+    1         2
+  /            \
+3                 4
+  \             /   \
+    5          6     7
+```
+
+You should convert it to:
+
+```graph
+     0
+  /     \
+5         4
+        /   \
+       6     7
+```

problems/tree/full-binary-tree-conversion/approach.js

@@ -0,0 +1,71 @@
+class TreeNode {
+  constructor(value) {
+    this.value = value;
+    this.left = null;
+    this.right = null;
+  }
+}
+
+// Method to convert the binary tree to a full binary tree
+function convertToFullBinaryTree(node) {
+  let tempL = null,
+    tempR = null;
+
+  // If the node is a leaf node, return it
+  if (node.left === null && node.right === null) {
+    return node;
+  }
+
+  // Recursively process the left and right subtrees
+  if (node.left !== null) {
+    tempL = convertToFullBinaryTree(node.left);
+  }
+  if (node.right !== null) {
+    tempR = convertToFullBinaryTree(node.right);
+  }
+
+  // If the node has two children, keep the node
+  if (node.left !== null && node.right !== null) {
+    node.left = tempL;
+    node.right = tempR;
+    return node;
+  }
+
+  // If the node has only one child, return the non-null child
+  return tempL === null ? tempR : tempL;
+}
+
+// Method to print the tree in InOrder (used for testing)
+function printInOrder(node) {
+  if (node === null) return;
+
+  printInOrder(node.left);
+  process.stdout.write(node.value + " ");
+  printInOrder(node.right);
+}
+
+// Main function to test the functionality
+function main() {
+  // Construct the binary tree
+  let root = new TreeNode(0);
+  root.left = new TreeNode(1);
+  root.right = new TreeNode(2);
+  root.left.left = new TreeNode(3);
+  root.left.left.right = new TreeNode(5);
+  root.right.right = new TreeNode(4);
+  root.right.right.left = new TreeNode(6);
+  root.right.right.right = new TreeNode(7);
+
+  console.log("Original Binary Tree (InOrder):");
+  printInOrder(root);
+  console.log();
+
+  // Convert to full binary tree
+  let fullTree = convertToFullBinaryTree(root);
+
+  console.log("Full Binary Tree (InOrder):");
+  printInOrder(fullTree);
+  console.log();
+}
+
+main();

problems/tree/full-binary-tree-conversion/approach.py

@@ -0,0 +1,63 @@
+class TreeNode:
+    def __init__(self, value):
+        self.value = value
+        self.left = None
+        self.right = None
+
+# Method to convert the binary tree to a full binary tree
+def convert_to_full_binary_tree(node):
+    tempL = None
+    tempR = None
+
+    # If the node is a leaf node, return it
+    if node.left is None and node.right is None:
+        return node
+
+    # Recursively process the left and right subtrees
+    if node.left is not None:
+        tempL = convert_to_full_binary_tree(node.left)
+    if node.right is not None:
+        tempR = convert_to_full_binary_tree(node.right)
+
+    # If the node has two children, keep the node
+    if node.left is not None and node.right is not None:
+        node.left = tempL
+        node.right = tempR
+        return node
+
+    # If the node has only one child, return the non-null child
+    return tempL if tempL is not None else tempR
+
+# Method to print the tree in InOrder (used for testing)
+def print_in_order(node):
+    if node is None:
+        return
+    print_in_order(node.left)
+    print(node.value, end=" ")
+    print_in_order(node.right)
+
+# Main function to test the functionality
+def main():
+    # Construct the binary tree
+    root = TreeNode(0)
+    root.left = TreeNode(1)
+    root.right = TreeNode(2)
+    root.left.left = TreeNode(3)
+    root.left.left.right = TreeNode(5)
+    root.right.right = TreeNode(4)
+    root.right.right.left = TreeNode(6)
+    root.right.right.right = TreeNode(7)
+
+    print("Original Binary Tree (InOrder):")
+    print_in_order(root)
+    print()
+
+    # Convert to full binary tree
+    full_tree = convert_to_full_binary_tree(root)
+
+    print("Full Binary Tree (InOrder):")
+    print_in_order(full_tree)
+    print()
+
+if __name__ == "__main__":
+    main()