Binary Tree Right Side View

mediumcodingAlgorithmsBinary TreesBFS~20 min

Asked at Google, Meta

Overview

Given the root of a binary tree, return the values of nodes visible from the right side, ordered from top to bottom.

Constraints

The number of nodes is in the range [0, 100].
-100 <= Node.val <= 100

Examples

rightSideView([1, 2, 3, null, 5, null, 4]);
// => [1, 3, 4]

rightSideView([1, null, 3]);
// => [1, 3]

Solution

Reveal solution

function rightSideView(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  while (queue.length) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === size - 1) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

Resources

Binary Tree Right Side View — LeetCode

binary-tree-right-side-view.js

Binary Tree Right Side View

mediumcodingAlgorithmsBinary Trees

Overview

Given the root of a binary tree, return the values of nodes visible from the right side, ordered from top to bottom.

Constraints

The number of nodes is in the range [0, 100].
-100 <= Node.val <= 100

Examples

rightSideView([1, 2, 3, null, 5, null, 4]);
// => [1, 3, 4]

rightSideView([1, null, 3]);
// => [1, 3]

Solution

Reveal solution

function rightSideView(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  while (queue.length) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === size - 1) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

Resources

Binary Tree Right Side View — LeetCode

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103 of 103 problems