מריה בוטוסוב

יש לעקוף עץ בינארי ללא רקורסיה. בכל צומת יש לבצע פונקציה שנותנים על ערך של צומת.

public static void traverse(BinaryTree tree, Action<BinaryTree> callback) {
    BinaryTree prev = null;
    BinaryTree curr = tree;
    while (curr != null) {
        BinaryTree next;
        if(prev == null || prev == curr.parent) { // from top to bottom
            if(curr.left != null)
                next = curr.left;
            else {
                callback(curr);
                next = curr.right != null ? curr.right : curr.parent;
            }					
        } else if (prev == curr.left) { //from left bottom
            callback(curr);
            next = curr.right != null ? curr.right : curr.parent;
        } else { //from right bottom
            next = curr.parent;
        }
        prev = curr;
        curr = next;
    }		
}
 
public class BinaryTree {
    public int value;
    public BinaryTree left;
    public BinaryTree right;
    public BinaryTree parent;
 
    public BinaryTree(int value) {
        this.value = value;
    }
 
    public BinaryTree(int value, BinaryTree parent) {
        this.value = value;
        this.parent = parent;
    }
}