Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
- initial thought and implementation was bfs, traverse right child
- bfs more suited as it pushes all right child instead of going right branch only
Solution
// from solutions
// https://leetcode.com/problems/binary-tree-right-side-view/solutions/382850/simple-javascript-bfs-solution-using-queue/
var rightSideView = function (root) {
const result = [];
const queue = [];
if (root === null) return result;
queue.push(root);
while (queue.length !== 0) {
let size = queue.length;
// console.log(queue)
for (let i = 0; i < size; i++) {
let n = queue.shift();
// only pushing, right node [left, right] in queue, or if it's the only node, e.g index 0
if (i === size - 1) result.push(n.val);
if (n.left !== null) queue.push(n.left);
if (n.right !== null) queue.push(n.right);
}
}
return result;
};
