257. Binary Tree Paths
EasyView on LeetCode
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Intuition
DFS with backtracking: build the path string as you go down. When you reach a leaf, add the complete path to results.
Algorithm
- 1DFS(node, path):
- 2If leaf: result.Add(path + node.val).
- 3Else: DFS(left, path + node.val + "->"); DFS(right, path + node.val + "->").
Example Walkthrough
Input: root = [1,2,3,null,5]
- 1.From 1->2->5: path="1->2->5". From 1->3: path="1->3".
Output: ["1->2->5","1->3"]
Common Pitfalls
- •Use string concatenation or StringBuilder carefully - in C# string is immutable so no explicit backtracking needed.
257.cs
C#
// Approach: DFS (preorder) with path tracking using a string builder.
// At each non-null node, append the node value (and '->' if not a leaf) to the current path.
// When a leaf node is reached, add the complete path string to the results list.
// Backtrack by passing the path string as a parameter (immutable strings make backtracking free).
// Alternatively, use a StringBuilder with explicit Remove calls if performance matters at scale.
// Time: O(n x L) where L is the average path length. Space: O(h) for the recursion stack.
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<string> BinaryTreePaths(TreeNode root)
{
List<string> ans = new List<string>();
Dfs(root, new StringBuilder(), ans);
return ans;
}
private void Dfs(TreeNode root, StringBuilder sb, List<string> ans)
{
if (root == null)
return;
if (root.left == null && root.right == null)
{
ans.Add(sb.Append(root.val).ToString());
return;
}
int length = sb.Length;
Dfs(root.left, sb.Append(root.val).Append("->"), ans);
sb.Length = length;
Dfs(root.right, sb.Append(root.val).Append("->"), ans);
sb.Length = length;
}
}Advertisement
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