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184 Chap. 5 Binary Trees/** Build a Huffman tree from list hufflist */static HuffTree buildTree() {HuffTree tmp1, tmp2, tmp3 = null;}while (Hheap.heapsize() > 1) { // While two items lefttmp1 = Hheap.removemin();tmp2 = Hheap.removemin();tmp3 = new HuffTree(tmp1.root(), tmp2.root(),tmp1.weight() + tmp2.weight());Hheap.insert(tmp3); // Return new tree to heap}return tmp3;// Return the treeFigure 5.29 Implementation for the Huffman tree construction function.buildHuff takes as input fl, the min-heap of partial Huffman trees, whichinitially are single leaf nodes as shown in Step 1 of Figure 5.25. The body offunction buildTree consists mainly of a for loop. On each iteration of thefor loop, the first two partial trees are taken off the heap and placed in variablestemp1 and temp2. A tree is created (temp3) such that the left and right subtreesare temp1 and temp2, respectively. Finally, temp3 is returned to fl.VUl1l2XFigure 5.30 An impossible Huffman tree, showing the situation where the twonodes with least weight, l 1 and l 2 , are not the deepest nodes in the tree. Trianglesrepresent subtrees.Proof: The proof is by induction on n, the number of letters.• Base Case: For n = 2, the Huffman tree must have the minimum externalpath weight because there are only two possible trees, each with identicalweighted path lengths for the two leaves.• Induction Hypothesis: Assume that any tree created by buildHuff thatcontains n − 1 leaves has minimum external path length.• Induction Step: Given a Huffman tree T built by buildHuff with nleaves, n ≥ 2, suppose that w 1 ≤ w 2 ≤ · · · ≤ w n where w 1 to w n arethe weights of the letters. Call V the parent of the letters with frequencies w 1and w 2 . From the lemma, we know that the leaf nodes containing the letterswith frequencies w 1 and w 2 are as deep as any nodes in T. If any other leaf