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    <p>First, I assume that your <code>level order traversal</code> is basically a BFS.</p> <p>Now, let's have a look at the string. Performing the BFS, we print "1" if the current node's got two sons. Otherwise, it's a leaf and we print 0, terminating the processing of the current branch.</p> <p>Consequently, during the reverse task, we can remember the list of open branches' last nodes and append the incoming nodes there.</p> <p>Let's demonstrate this approach on an example:</p> <pre><code>Level 1: Tree : 1 - id 0 Open branches : 0 0 (left and right son) Remaining string : 11001000 ********* Level 2: Tree : 1 1 1 Open branches : 1 1 2 2 Remaining string : 001000 ********* Level 3: Tree : 1 1 1 0 0 1 0 Open branches : 5 5 Remaining string : 00 Level 4: Tree : 1 1 1 0 0 1 0 0 0 No more input, we're done. </code></pre> <p>Having the tree, the post-order traversal is trivial.</p> <p>And the code (it assumes that the tree is quite dense, otherwise it's not very memory efficient): </p> <pre><code>import java.util.ArrayDeque; import java.util.Queue; public class Main { static final int MAX_CONST = 50; public static void main(String[] args) { String evilString = "111001000"; // Assuming this string is a correct input char[] treeRepr = new char[MAX_CONST]; Queue&lt;Integer&gt; q = new ArrayDeque&lt;Integer&gt;(); q.add(0); for (int i = 0; i &lt; evilString.length(); ++i) { int index = q.remove(); char ch = evilString.charAt(i); if (ch == '1') { q.add(2*(index+1)-1); q.add(2*(index+1)); } treeRepr[index] = ch; // System.out.println(q.size()); } System.out.println(arrToString(treeRepr, 0, new StringBuilder())); } public static StringBuilder arrToString(char[] array, int index, StringBuilder sb) { if (array[index] == '1') { arrToString(array, 2*(index+1)-1, sb); arrToString(array, 2*(index+1), sb); } sb.append(array[index]); return sb; } } </code></pre>
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    1. POHow to construct a binary tree from just the level order traversal string
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    1. USDanstahr
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