Pre-Order and Post-Order Traversals in the same Problem

Very interesting problem: https://leetcode.com/problems/boundary-of-binary-tree/

545. Boundary of Binary Tree

Medium

Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes.  (The values of the nodes may still be duplicates.)

Left boundary is defined as the path from root to the left-most node. Right boundary is defined as the path from root to the right-most node. If the root doesn't have left subtree or right subtree, then the root itself is left boundary or right boundary. Note this definition only applies to the input binary tree, and not applies to any subtrees.

The left-most node is defined as a leaf node you could reach when you always firstly travel to the left subtree if exists. If not, travel to the right subtree. Repeat until you reach a leaf node.

The right-most node is also defined by the same way with left and right exchanged.

Example 1

Input:
  1
   \
    2
   / \
  3   4

Ouput:
[1, 3, 4, 2]

Explanation:
The root doesn't have left subtree, so the root itself is left boundary.
The leaves are node 3 and 4.
The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed right boundary.
So order them in anti-clockwise without duplicates and we have [1,3,4,2].

Example 2

Input:
    ____1_____
   /          \
  2            3
 / \          / 
4   5        6   
   / \      / \
  7   8    9  10  
       
Ouput:
[1,2,4,7,8,9,10,6,3]

Explanation:
The left boundary are node 1,2,4. (4 is the left-most node according to definition)
The leaves are node 4,7,8,9,10.
The right boundary are node 1,3,6,10. (10 is the right-most node).
So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].

The idea here would be to break the problem apart:
1) Deal with the special cases first (null, root = leaf, etc.). Easy
2) Create 3 functions: one that gets all the leaves using pre-order traversal, one that gets the left boundary also using pre-order and one that gets the right boundary using post-order
3) Some logic is necessary to avoid adding the root more than once
4) Merge all the lists
5) Voila!

Works quite well and fast. Cheers, stay home, stay safe!!! ACC

public class Solution
{
    public IList<int> BoundaryOfBinaryTree(TreeNode root)
    {
        //Special cases first: (1) root is null, (2) root is leaf
        if (root == null) return new List<int>();
        if (root.left == null && root.right == null) return new List<int> { root.val };

        //All cases:
        List<int> retVal = new List<int> { root.val };

        List<int> leaves = new List<int>();
        FindLeavesPreOrder(root, leaves);
        List<int> left = new List<int>();
        bool foundFirstLeaf = false;
        if (root.left != null)
        {
            FindLeftPath(root, left, ref foundFirstLeaf);
            if (left != null && left.Count > 0) left.RemoveAt(0); //Remember: root already added!
        }

        List<int> right = new List<int>();
        foundFirstLeaf = false;
        if (root.right != null)
        {
            FindRightPath(root, right, ref foundFirstLeaf);
            if (right != null && right.Count > 0) right.RemoveAt(right.Count - 1); //Remember: root already added!
        }

        retVal.AddRange(left);
        retVal.AddRange(leaves);
        retVal.AddRange(right);

        return retVal;
    }

    private void FindRightPath(TreeNode node, IList<int> right, ref bool foundFirstLeaf)
    {
        if (node == null || foundFirstLeaf) return;

        if (node.left == null && node.right == null)
        {
            foundFirstLeaf = true;
            return;
        }

        FindRightPath(node.right, right, ref foundFirstLeaf);
        FindRightPath(node.left, right, ref foundFirstLeaf);

        right.Add(node.val);
    }

    private void FindLeftPath(TreeNode node, IList<int> left, ref bool foundFirstLeaf)
    {
        if (node == null || foundFirstLeaf) return;

        if (node.left == null && node.right == null)
        {
            foundFirstLeaf = true;
            return;
        }

        left.Add(node.val);

        FindLeftPath(node.left, left, ref foundFirstLeaf);
        FindLeftPath(node.right, left, ref foundFirstLeaf);
    }

    private void FindLeavesPreOrder(TreeNode node, IList<int> leaves)
    {
        if (node == null) return;
        if (node.left == null && node.right == null) leaves.Add(node.val);
        FindLeavesPreOrder(node.left, leaves);
        FindLeavesPreOrder(node.right, leaves);
    }
}