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199.Binary-Tree-Right-Side-View.md

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199. Binary Tree Right Side View

Problem

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.

Example 1:

https://assets.leetcode.com/uploads/2021/02/14/tree.jpg

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

Input: root = [1,null,3]
Output: [1,3]

Example 3:

Input: root = []
Output: []

Constraints:

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

Note

Code

  • PHP

    Runtime 3ms Beats 88.14% of users with PHP

    Memory 19.75MB Beats 84.75% of users with PHP

    /*
 * Definition for a binary tree node.
 * class TreeNode {
 *     public $val = null;
 *     public $left = null;
 *     public $right = null;
 *     function __construct($val = 0, $left = null, $right = null) {
 *         $this->val = $val;
 *         $this->left = $left;
 *         $this->right = $right;
 *     }
 * }
 */
class Solution {

    /**
     * @param TreeNode $root
     * @return Integer[]
     */
    function rightSideView($root) {
        $result = [];
        $queue = [$root];

        while (!empty($queue)) {
            $tmpQueue = [];
            for ($i = 0; $i < count($queue); $i++) {
                $node = $queue[$i];
                if ($i == 0) {
                    $result[] = $node->val;
                }

                if (!is_null($node->right)) {
                    $tmpQueue[] = $node->right;
                }

                if (!is_null($node->left)) {
                    $tmpQueue[] = $node->left;
                }
            }

            $queue = $tmpQueue;
        }

        return $result;
    }
}

  • GoLang

    Runtime 0ms Beats 100.00% of users with Go

    Memory 2.52MB Beats **25.46%**of users with Go

    /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func rightSideView(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }

    result := []int{}
    queue := []*TreeNode{root}
    
    for len(queue) != 0 {
        newQueue := []*TreeNode{}
        for i := 0; i < len(queue); i++ {
            node := queue[i]

            if i == 0 {
                result = append(result, node.Val)
            }

            if node.Right != nil {
                newQueue = append(newQueue, node.Right)
            }

            if node.Left != nil {
                newQueue = append(newQueue, node.Left)
            }
        }

        queue = newQueue
    }

    return result
}

Reference