Closest Leaf in a Binary Tree

Problem

Given a binary tree where every node has a unique value, and a target key k, find the value of the nearest leaf node to target k in the tree.

Here, nearest to a leaf means the least number of edges travelled on the binary tree to reach any leaf of the tree. Also, a node is called a leaf if it has no children.

In the following examples, the input tree is represented in flattened form row by row. The actual root tree given will be a TreeNode object.

Examples

Example 1

Input:
root = [1, 3, 2], k = 1
Diagram of binary tree:
          1
         / \
        3   2

Output: 2 (or 3)

Explanation: Either 2 or 3 is the nearest leaf node to the target of 1.

Example 2

Input:
root = [1], k = 1
Output: 1

Explanation: The nearest leaf node is the root node itself.

Example 3

root = [1,2,3,4,null,null,null,5,null,6], k = 2
Diagram of binary tree:
             1
            / \
           2   3
          /
         4
        /
       5
      /
     6

Output: 3
Explanation: The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.

Constraints

1. root represents a binary tree with at least 1 node and at most 1000 nodes.
2. Every node has a unique node.val in range [1, 1000].
3. There exists some node in the given binary tree for which node.val == k.

Solution

Method 1 - DFS

Reasoning and Approach

To find the nearest leaf node to the target k, the solution requires the following steps:

1. Convert the Input Tree into a Graph:
   • Each node in the binary tree is treated as a graph node, connecting its parent and children. This essentially transforms the tree into an undirected graph.
   • By doing so, we can perform a Breadth-First Search (BFS) from the target node k to find the nearest leaf.
2. Locate the Target Node:
   • Traverse the binary tree while constructing the graph to locate the target node k.
3. Perform BFS from Target Node:
   • Starting from the node k, perform BFS to find the nearest leaf node (i.e., a node without children).

Code

Java

class Solution {
    static class TreeNode {
        int val;
        TreeNode left, right;
        TreeNode(int x) { val = x; }
    }
    
    public int findClosestLeaf(TreeNode root, int k) {
        // Map to hold graph connections
        Map<TreeNode, List<TreeNode>> graph = new HashMap<>();
        // Locate the target node
        TreeNode target = buildGraph(root, null, graph, k);

        // BFS to find the closest leaf
        Queue<TreeNode> q = new LinkedList<>();
        Set<TreeNode> visited = new HashSet<>();
        q.add(target);
        visited.add(target);
        
        while (!q.isEmpty()) {
            TreeNode curr = q.poll();
            // Check if current node is a leaf node
            if (curr.left == null && curr.right == null) {
                return curr.val;
            }
            // Explore neighbours
            for (TreeNode neighbour : graph.get(curr)) {
                if (!visited.contains(neighbour)) {
                    visited.add(neighbour);
                    q.add(neighbour);
                }
            }
        }
        return -1; // Should not reach here for valid input
    }
    
    private TreeNode buildGraph(TreeNode node, TreeNode parent, Map<TreeNode, List<TreeNode>> graph, int k) {
        if (node == null) return null;
        // Create connections
        graph.putIfAbsent(node, new ArrayList<>());
        if (parent != null) {
            graph.get(node).add(parent);
            graph.get(parent).add(node);
        }
        // Search for the target node
        if (node.val == k) return node;
        TreeNode left = buildGraph(node.left, node, graph, k);
        TreeNode right = buildGraph(node.right, node, graph, k);
        return left != null ? left : right;
    }
}

Python

class TreeNode:
    def __init__(self, val: int = 0, left: Optional['TreeNode'] = None, right: Optional['TreeNode'] = None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def findClosestLeaf(self, root: TreeNode, k: int) -> int:
        # Dictionary to store graph connections
        graph: dict[TreeNode, list[TreeNode]] = {}
        
        # Build graph and locate target node
        def dfs(node: Optional[TreeNode], parent: Optional[TreeNode]) -> Optional[TreeNode]:
            if not node:
                return None
            if node not in graph:
                graph[node] = []
            if parent:
                graph[node].append(parent)
                graph[parent].append(node)
            if node.val == k:
                return node
            left = dfs(node.left, node)
            right = dfs(node.right, node)
            return left or right
        
        target = dfs(root, None)
        
        # BFS to find the nearest leaf
        queue: deque[TreeNode] = deque([target])
        visited: set[TreeNode] = {target}
        
        while queue:
            curr = queue.popleft()
            # Check if current node is a leaf
            if not curr.left and not curr.right:
                return curr.val
            # Traverse neighbours
            for neighbour in graph[curr]:
                if neighbour not in visited:
                    visited.add(neighbour)
                    queue.append(neighbour)
        
        return -1  # Should not reach here for valid input

Complexity

• ⏰ Time complexity: O(n)
  • Graph Construction: The dfs traversal constructs the graph in O(n) where n is the number of nodes in the binary tree.
  • BFS Search: The BFS traversal visits each node at most once to locate the nearest leaf, taking O(n).
  • Overall Complexity: O(n) + O(n) = O(n).
• 🧺 Space complexity: O(n)
  • Graph Storage: The graph uses O(n) extra space to store connections between nodes.
  • BFS Queue and Visited Set: Both the queue and set require O(n) space in the worst case.
  • Overall Complexity: O(n).