Binary Tree Paths

Easy Recursion & Backtracking Basic Recursion Original on LeetCode

Binary Tree Paths is a easy recursion & backtracking problem solved with the basic recursion pattern. The best approach, backtracking with one shared stringbuilder, runs in O(n · h) time and O(h) space. Below are 2 approaches in Java, from dfs building new strings up.

Problem

Return every root-to-leaf path in a binary tree, formatted like "1->2->5", in any order.

Examples

Example 1

Input: root = [1, 2, 3, null, 5]
Output: ["1->2->5", "1->3"]

Example 2

Input: root = [7]
Output: ["7"]

Constraints

The tree has 1 to 100 nodes.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
DFS building new strings	`O(n · h)`	`O(n · h)`
Backtracking with one shared StringBuilder	`O(n · h)`	`O(h)`

1DFS building new strings

TimeO(n · h)Each string copy costs up to h characters.

SpaceO(n · h)

Pass the path as a string. Each call makes a new string, so nothing needs undoing.

path += node.val; at a leaf, add path.
Otherwise recurse with path + "->".

Java

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> out = new ArrayList<>();
        dfs(root, "", out);
        return out;
    }

    private void dfs(TreeNode node, String path, List<String> out) {
        if (node == null) return;
        path += node.val;
        if (node.left == null && node.right == null) { out.add(path); return; }
        dfs(node.left, path + "->", out);
        dfs(node.right, path + "->", out);
    }
}

2Backtracking with one shared StringBuilder

TimeO(n · h)Copying a path into the answer still costs O(h) per leaf.

SpaceO(h)Recursion depth plus one builder, not counting the output.

Append to a single StringBuilder on the way down and cut it back to its previous length on the way up. This is the choose / explore / un-choose template on a tree.

len = sb.length(); append val (and '->' when not the root).
At a leaf, record sb.toString().
Recurse left and right, then sb.setLength(len).

Java

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> out = new ArrayList<>();
        dfs(root, new StringBuilder(), out);
        return out;
    }

    private void dfs(TreeNode node, StringBuilder sb, List<String> out) {
        if (node == null) return;
        int len = sb.length();
        if (len > 0) sb.append("->");
        sb.append(node.val);
        if (node.left == null && node.right == null) out.add(sb.toString());
        else {
            dfs(node.left, sb, out);
            dfs(node.right, sb, out);
        }
        sb.setLength(len);
    }
}

Edge cases to test

Single node
Negative values
A node with only one child is not a leaf

Hints

Hint 1

Carry the path down the tree; record it only at a leaf (no children).

FAQ

What is the best time complexity for Binary Tree Paths?

Backtracking with one shared StringBuilder runs in O(n · h) time and O(h) extra space. Copying a path into the answer still costs O(h) per leaf.

Which pattern does Binary Tree Paths use?

It is a recursion & backtracking problem that uses the basic recursion pattern. Other problems with the same pattern: Factorial of a number, Fibonacci Number, Binary Tree Inorder Traversal (Recursive).

Is there a brute force solution for Binary Tree Paths?

Yes. DFS building new strings takes O(n · h) time and O(n · h) space. Pass the path as a string.

Which edge cases should I test for Binary Tree Paths?

Single node; Negative values; A node with only one child is not a leaf.