<pre class="shiki github-dark shaku"><code>// This is the class for the node// you can use this directly as it is bundled with your code// class Node {// value: number// left: null | Node// right: null | Node// constructor(val) {// this.value = val// this.left = null// this.right = null// }// }/** * @param {Node} root * @returns {number[]} */function traverse(root) { // your code here // Level order traversal, assign row and column // Final output =&gt; column then row // Map&lt;column, [[minRow, maxRow], &lt;Map&lt;row, [node]&gt;&gt; const map = new Map(); const queue = []; let startingCol = 0; queue.push([root, 0, startingCol]); while (queue.length) { const n = queue.length; for (let i = 0; i &lt; n; i++) { const [node, row, col] = queue.shift(); if (map.has(col)) { const [rowRange, rowMap] = map.get(col); rowRange[0] = Math.min(rowRange[0], row); rowRange[1] = Math.max(rowRange[1], row); if (rowMap.has(row)) { rowMap.get(row).push(node); } else { rowMap.set(row, [node]); } } else { map.set(col, [[row, row], new Map([[row, [node]]])]); }			 if (node.left) { 	startingCol = Math.min(startingCol, col-1); queue.push([node.left, row+1, col-1, col]); } if (node.right) { queue.push([node.right, row+1, col+1, col]); } } // If next queue has nodes with same row and col, need to re-arrange base on parent's column order queue.sort(([a, aRow, aCol, aParentCol],[b, bRow, bCol, bParentCol]) =&gt; { 	if (aRow === bRow &amp;&amp; aCol === bCol) { return aParentCol &lt; bParentCol ? -1 : 1; } return 0; }) } const output = []; for (let i = startingCol; i &lt; startingCol + map.size; i++) { const [[lo, hi], rowMap] = map.get(i); for (let j = lo; j &lt;= hi; j++) { if (rowMap.has(j)) { output.push(...rowMap.get(j)); } } } return output.map(node =&gt; node.value);}//const tree = new Node(1);//tree.left = new Node(2);//tree.right = new Node(3);//tree.left.right = new Node(4);//tree.right.left = new Node(5);//tree.right.left.left = new Node(6);//tree.left.right.right = new Node(7);//tree.right.left.left.right = new Node(8);//tree.right.left.left.right.right = new Node(11);//tree.left.right.right.left = new Node(9);//tree.left.right.right.left.left = new Node(10);//console.log(traverse(tree));</code></pre>

Level order traversal + 2D Map.

Traverse a binary tree vertically from left to right, from top to bottom, the binary tree contains positive integers as node values. [https://cdn.bfe.…