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  1. PO
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    <p>Looks like a straightforward algorithm: Find the root of the search tree in the game tree and check whether the children of the search tree are a subset of the children in the game tree.</p> <p>From your explanations, I'm not sure whether the search tree</p> <pre><code> A / \ A A </code></pre> <p>should match this tree:</p> <pre><code> A /|\ A C A </code></pre> <p>(i.e. if non-matching children are supposed to be ignored.)</p> <p>Anyway, here's the code I just toyed around with. It's a fully running example and comes with a main method and a simple <code>Node</code> class. Feel free to play with it:</p> <pre><code>import java.util.Vector; public class PartialTreeMatch { public static void main(String[] args) { Node testTree = createTestTree(); Node searchTree = createSearchTree(); System.out.println(testTree); System.out.println(searchTree); partialMatch(testTree, searchTree); } private static boolean partialMatch(Node tree, Node searchTree) { Node subTree = findSubTreeInTree(tree, searchTree); if (subTree != null) { System.out.println("Found: " + subTree); return true; } return false; } private static Node findSubTreeInTree(Node tree, Node node) { if (tree.value == node.value) { if (matchChildren(tree, node)) { return tree; } } Node result = null; for (Node child : tree.children) { result = findSubTreeInTree(child, node); if (result != null) { if (matchChildren(tree, result)) { return result; } } } return result; } private static boolean matchChildren(Node tree, Node searchTree) { if (tree.value != searchTree.value) { return false; } if (tree.children.size() &lt; searchTree.children.size()) { return false; } boolean result = true; int treeChildrenIndex = 0; for (int searchChildrenIndex = 0; searchChildrenIndex &lt; searchTree.children.size(); searchChildrenIndex++) { // Skip non-matching children in the tree. while (treeChildrenIndex &lt; tree.children.size() &amp;&amp; !(result = matchChildren(tree.children.get(treeChildrenIndex), searchTree.children.get(searchChildrenIndex)))) { treeChildrenIndex++; } if (!result) { return result; } } return result; } private static Node createTestTree() { Node subTree1 = new Node('A'); subTree1.children.add(new Node('A')); subTree1.children.add(new Node('A')); Node subTree2 = new Node('A'); subTree2.children.add(new Node('A')); subTree2.children.add(new Node('C')); subTree2.children.add(subTree1); Node subTree3 = new Node('B'); subTree3.children.add(subTree2); Node root = new Node('A'); root.children.add(subTree3); return root; } private static Node createSearchTree() { Node root = new Node('A'); root.children.add(new Node('A')); root.children.add(new Node('A')); return root; } } class Node { char value; Vector&lt;Node&gt; children; public Node(char val) { value = val; children = new Vector&lt;Node&gt;(); } public String toString() { StringBuilder sb = new StringBuilder(); sb.append('('); sb.append(value); for (Node child : children) { sb.append(' '); sb.append(child.toString()); } sb.append(')'); return sb.toString(); } } </code></pre>
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    1. POEasy way to find Subtree in a Tree
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    1. PTAnswer
    1. USgclj5
    1. USgclj5
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    1. This table or related slice is empty.
    1. POEasy way to find Subtree in a Tree
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    1. VO
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      1. VTUpMod
    2. VO
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      1. VTUpMod
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      1. VTAcceptedByOriginator
    1. COBetter edit that before the 'Vector' police get you! ;-)
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      1. USOutlaw Programmer
    2. COLet them come! :)
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      1. USgclj5
 

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