Find Leaves of Binary Tree

LeetCode Link

Problem Description

Visit LeetCode for the full problem description.

---

Solutions

Solution 1: C# (Best: 196 ms)

Metric	Value
Runtime	196 ms
Memory	41 MB
Date	2022-01-31

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public IList<IList<int>> FindLeaves(TreeNode root) {
        List<IList<int>> result = new List<IList<int>>();
    Height(root, result);
    return result;
}

private int Height(TreeNode root, List<IList<int>> result)
{
    if(root==null)
    {
        return -1;
    }
    int height = 1 + Math.Max(Height(root.left, result), Height(root.right, result));
    while(result.Count <= height)
    {
        result.Add(new List<int>());
    }
    result[height].Add(root.val);
    root.left = null;
    root.right = null;
    return height;  
}
    }

📜 1 more C# submission(s)

Submission (2022-01-31) — 201 ms, 41.3 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public IList<IList<int>> FindLeaves(TreeNode root) {
        List<IList<int>> result = new List<IList<int>>();
    Height(root, result);
    return result;
}

private int Height(TreeNode root, List<IList<int>> result)
{
    if(root==null)
    {
        return -1;
    }
    int height = 1 + Math.Max(Height(root.left, result), Height(root.right, result));
    while(result.Count < height+1)
    {
        result.Add(new List<int>());
    }
    result[height].Add(root.val);
    root.left = null;
    root.right = null;
    return height;  
}
    }

---

Complexity Analysis

Approach	Time	Space
Solution	To be analyzed	To be analyzed