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].

[Problem Link]

The solution to this problem can be constructed using 3 sets:-

  • Left boundary
  • Leaf Nodes (left to right)
  • Right boundary

Generating these sets is straightforward but we need to be mindful of double counting nodes.

  • Left boundary and right boundary will contain root so best to separate that out.
  • Left boundary’s leftmost node will also be a leaf node (same for right boundary’s rightmost node). Again, best to remove these from those sets.

Our solution will be concatenation of

  • Root Node
  • Left Boundary Nodes (without the last leaf node)
  • Leaf nodes (from left to right)
  • In reverse, Right Boundary Nodes (without the last leaf node)
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
 
// get all leaf nodes (left to right)
var preOrder = function(root, level, leafs) {
    if (!root) return;
    if (!root.left && !root.right && level !== 0) leafs.push(root.val);    
    preOrder(root.left, level+1, leafs);
    preOrder(root.right, level+1, leafs);
}
 
var boundaryOfBinaryTree = function(root) {
 
    if (!root) return [];
 
    // get all left boundary nodes
    let leftList = [];
    let curr = root.left;
    while (curr) {
        leftList.push(curr.val);
        if (curr.left) curr = curr.left;
        else curr = curr.right;
    }
    // without the leftmost leaf (because that will already be part of the leaf's list)
    leftList.pop();
 
    // get all right boundary nodes
    let rightList = [];
    curr = root.right;
    while (curr) {
        rightList.push(curr.val);
        if (curr.right) curr = curr.right;
        else curr = curr.left;
    }
    // without the rightmost leaf (because that will already be part of the leaf's list)
    rightList.pop();
 
    // get all leaf nodes
    let leafList = [];
    preOrder(root, 0, leafList);
 
    // Solution is:
    // Root All_Left_Nodes All_Leaf_Nodes Reverse(All_Right_Nodes)
    // Remember to remove last leaf node from All_Left_Nodes and All_Right_Nodes otherwise you will double count!
 
    return ([root.val]).
            concat(leftList).
            concat(leafList).
            concat(rightList.reverse());
};