Use 0-based array indexing
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@ -774,21 +774,21 @@ The following array corresponds to the above tree:
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\node at (14.5,0.5) {$1$};
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\node at (14.5,0.5) {$1$};
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\footnotesize
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\footnotesize
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\node at (0.5,2.5) {$1$};
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\node at (0.5,2.5) {$0$};
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\node at (1.5,2.5) {$2$};
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\node at (1.5,2.5) {$1$};
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\node at (2.5,2.5) {$3$};
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\node at (2.5,2.5) {$2$};
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\node at (3.5,2.5) {$4$};
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\node at (3.5,2.5) {$3$};
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\node at (4.5,2.5) {$5$};
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\node at (4.5,2.5) {$4$};
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\node at (5.5,2.5) {$6$};
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\node at (5.5,2.5) {$5$};
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\node at (6.5,2.5) {$7$};
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\node at (6.5,2.5) {$6$};
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\node at (7.5,2.5) {$8$};
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\node at (7.5,2.5) {$7$};
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\node at (8.5,2.5) {$9$};
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\node at (8.5,2.5) {$8$};
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\node at (9.5,2.5) {$10$};
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\node at (9.5,2.5) {$9$};
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\node at (10.5,2.5) {$11$};
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\node at (10.5,2.5) {$10$};
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\node at (11.5,2.5) {$12$};
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\node at (11.5,2.5) {$11$};
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\node at (12.5,2.5) {$13$};
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\node at (12.5,2.5) {$12$};
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\node at (13.5,2.5) {$14$};
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\node at (13.5,2.5) {$13$};
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\node at (14.5,2.5) {$15$};
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\node at (14.5,2.5) {$14$};
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\end{tikzpicture}
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\end{tikzpicture}
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\end{center}
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\end{center}
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@ -846,27 +846,27 @@ can be found as follows:
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\node at (14.5,0.5) {$1$};
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\node at (14.5,0.5) {$1$};
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\footnotesize
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\footnotesize
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\node at (0.5,2.5) {$1$};
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\node at (0.5,2.5) {$0$};
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\node at (1.5,2.5) {$2$};
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\node at (1.5,2.5) {$1$};
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\node at (2.5,2.5) {$3$};
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\node at (2.5,2.5) {$2$};
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\node at (3.5,2.5) {$4$};
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\node at (3.5,2.5) {$3$};
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\node at (4.5,2.5) {$5$};
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\node at (4.5,2.5) {$4$};
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\node at (5.5,2.5) {$6$};
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\node at (5.5,2.5) {$5$};
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\node at (6.5,2.5) {$7$};
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\node at (6.5,2.5) {$6$};
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\node at (7.5,2.5) {$8$};
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\node at (7.5,2.5) {$7$};
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\node at (8.5,2.5) {$9$};
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\node at (8.5,2.5) {$8$};
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\node at (9.5,2.5) {$10$};
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\node at (9.5,2.5) {$9$};
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\node at (10.5,2.5) {$11$};
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\node at (10.5,2.5) {$10$};
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\node at (11.5,2.5) {$12$};
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\node at (11.5,2.5) {$11$};
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\node at (12.5,2.5) {$13$};
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\node at (12.5,2.5) {$12$};
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\node at (13.5,2.5) {$14$};
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\node at (13.5,2.5) {$13$};
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\node at (14.5,2.5) {$15$};
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\node at (14.5,2.5) {$14$};
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\end{tikzpicture}
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\end{tikzpicture}
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\end{center}
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\end{center}
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Node 5 is at position 3, node 8 is at position 6,
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Node 5 is at position 2, node 8 is at position 5,
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and the node with lowest level between
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and the node with lowest level between
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positions $3 \ldots 6$ is node 2 at position 4
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positions $2 \ldots 5$ is node 2 at position 3
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whose level is 2.
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whose level is 2.
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Thus, the lowest common ancestor of
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Thus, the lowest common ancestor of
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nodes 5 and 8 is node 2.
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nodes 5 and 8 is node 2.
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