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Add inorder binary tree traversal (#3898)
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package com.thealgorithms.datastructures.trees;
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import java.util.ArrayDeque;
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import java.util.ArrayList;
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import java.util.Deque;
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import java.util.List;
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/**
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* Given tree is traversed in an 'inorder' way: LEFT -> ROOT -> RIGHT.
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* Below are given the recursive and iterative implementations.
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*
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* Complexities:
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* Recursive: O(n) - time, O(n) - space, where 'n' is the number of nodes in a tree.
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*
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* Iterative: O(n) - time, O(h) - space, where 'n' is the number of nodes in a tree
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* and 'h' is the height of a binary tree.
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* In the worst case 'h' can be O(n) if tree is completely unbalanced, for instance:
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* 5
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* \
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* 6
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* \
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* 7
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* \
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* 8
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*
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* @author Albina Gimaletdinova on 21/02/2023
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*/
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public class InorderTraversal {
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public static List<Integer> recursiveInorder(BinaryTree.Node root) {
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List<Integer> result = new ArrayList<>();
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recursiveInorder(root, result);
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return result;
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}
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public static List<Integer> iterativeInorder(BinaryTree.Node root) {
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List<Integer> result = new ArrayList<>();
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if (root == null) return result;
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Deque<BinaryTree.Node> stack = new ArrayDeque<>();
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while (!stack.isEmpty() || root != null) {
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while (root != null) {
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stack.push(root);
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root = root.left;
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}
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root = stack.pop();
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result.add(root.data);
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root = root.right;
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}
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return result;
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}
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private static void recursiveInorder(BinaryTree.Node root, List<Integer> result) {
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if (root == null) {
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return;
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}
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recursiveInorder(root.left, result);
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result.add(root.data);
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recursiveInorder(root.right, result);
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}
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}
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