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    <p>The simple answer is.. Just change the queue to a stack! (assuming it's a binary tree)</p> <p>If you want to traverse from left to right you should invert the order of the pushes (like I did):</p> <pre><code>stack.push(rootNode); while(stack.Any()) { var currentNode = stack.pop(); if(currentNode.data == searchedData) break; if(currentNode.Right != null) stack.push(currentNode.Right); if(currentNode.Left != null) stack.push(currentNode.Left); } </code></pre> <p>So.. This will traverse the tree:</p> <p><img src="https://i.stack.imgur.com/ewxAr.jpg" alt="enter image description here"></p> <p>in the order:</p> <p>A -> B -> D -> E -> C -> F -> G</p> <p>The order of operations will be:</p> <ol> <li>push(a)</li> <li>pop ; a</li> <li>push(c)</li> <li>push(b)</li> <li>pop ; b</li> <li>push(e)</li> <li>push(d)</li> <li>pop ; d</li> <li>pop ; e</li> <li>pop ; c</li> <li>push(g)</li> <li>push(f)</li> <li>pop f</li> <li>pop g</li> </ol> <h1>Another way</h1> <p>There's a similar way using recursion (it will be like using an implicit stack)</p> <pre class="lang-py prettyprint-override"><code>def dfsSearch(node, searchedData): # Base case 1: end of the tree if node == None: return None # Base case 2: data found if node.data == searchedData: return node # Recursive step1: Traverse left child left = dfsSearch(node.left, searchedData) if left != None: return left # Recursive step2: Traverse right child right = dfsSearch(node.right, searchedData) if right != None: return right return None </code></pre>
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    1. COthank you! I can't believe it was that easy.
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