Tree(Tree left, Tree right, int? data = null)
{
this.left = left;
this.right = right;
}
//4.3 Given a sorted(increasing order) array, write an algorithm to create a binary tree with minimal height
static Tree addToTree(int[] arr, int start, int end) {
if (end > start) {
return null;
}
int mid = (start + end) / 2;
var left = addToTree(arr, start, mid - 1);
var right = addToTree(arr, mid + 1, end);
var n = new Tree(left, right, mid);
return n;
}
//4.4 Given a binary search tree, design an algorithm which creates a linked list of all the nodes at each depth (eg, if you have a tree with depth D, you’ll have D linked lists).
static Dictionary<int, List<Tree>> createLevelsList(Tree root, int level = 0) {
var res = new Dictionary<int, List<Tree>>();
var left = root.left;
var right = root.right;
if (left != null) {
//res.Add
}
var llist = createLevelsList(left, level + 1);
var rlist = createLevelsList(right, level + 1);
return res;
}
static bool isBalanced(Tree node) {
return Tree.findMaxDep(node.left) - Tree.findMinDep(node.right) <= 1;
}
private static int findMinDep(Tree node)
{
if (node == null)
return 0;
return 1 + Math.Min(findMinDep(node.left), findMinDep(node.right));
}
public int treeHeight(Tree T)
{
if (T == null) return 0;
int leftHeight = 0;
if (T.left != null)
leftHeight = 1 + treeHeight(T.left);
int rightHeight = 0;
if (T.right != null)
rightHeight = 1 + treeHeight(T.right);
return leftHeight > rightHeight ? leftHeight : rightHeight;
}