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We showed how a doubly linked list helps here so what

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Unformatted text preview: . So what happens if we do keep a parent pointer, as well as left and right child pointers, in each node. The first node in inorder traversal is the left-most leftnode. struct node { void* data; curr = root; while (curr->left != 0) (currcurr = curr->left; curr- nodePtr parent, left, right; parent left right; }; All the insertion and deletion code would now be rewritten to update the parent pointer as well as parent the child pointers. But traversal is now very simple. if (curr->right != 0) // into right subtree (curr{ curr = curr->right; currwhile (curr->left != 0) (currcurr = curr->left; curr} If it If it has no right subtree and no parent, we're at the root of a tree with only a left subtree so we're finished. else if (curr->parent == 0) (currcurr = 0; Now let's suppose that at some time during the traversal we are at node curr. curr If it has a right child, we find the left-most node in leftthe right subtree. Otherwise the current node is the rightmost node of subtree. We look for an ancestor who has t...
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