# Exercise 10.4.5

$\star$ Write an $\O(n)$-time nonrecursive procedure that, given an $n$-node binary tree prints out the key of each node. Use no more than constant extra space outside the tree itself and do not modify the tree, even temporarily, during the procedure.

This is tricky. We need a pointer to the parent. We keep track of the previous pointer (starting with NIL) and do the following.

1. If we're coming from the parent, move to the left child
2. If we're coming from the left child, move to the right child
3. If we're coming from the right child, move to the parent

To handle cases of less than two children, we skip to the next step if the conditions allow for it. That is:

• If there is only a right child, and we're coming form the parent, we move to the right child
• If we come from the left child, but there is no right child, we move to the parent.
• If we there are no children, we move to the parent.

### C code

struct tree_t {
struct tree_t *left;
struct tree_t *right;
struct tree_t *parent;
int key;
};
typedef struct tree_t tree_t;
void store(int);

void print_tree(tree_t *tree) {
tree_t *prev;
prev = 0;

while (tree) {
if (prev == tree->parent) {
store(tree->key);
prev = tree;
tree = tree->left  ? tree->left :
tree->right ? tree->right :
tree->parent;
} else if (prev == tree->left && tree->right) {
prev = tree;
tree = tree->right;
} else {
prev = tree;
tree = tree->parent;
}
}
}

#define MAX_SIZE 10
int keys[MAX_SIZE];
int count = 0;

void reset_storage() {
count = 0;
}

void store(int key) {
keys[count++] = key;
}