742. Closest Leaf in a Binary Tree

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.
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:
Input:
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.

Note:

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.








My solution is simple:

1. First, preform DFS on root in order to find the node whose val = k, at the meantime use HashMap to keep record of all back edges from child to parent;
2. Then perform BFS on this node to find the closest leaf node.

class Solution {
    public int findClosestLeaf(TreeNode root, int k) {
        Map<TreeNode, List<TreeNode>> graph = new HashMap();
        dfs(graph, root, null);

        Queue<TreeNode> queue = new LinkedList();
        Set<TreeNode> seen = new HashSet();

        for (TreeNode node: graph.keySet()) {
            if (node != null && node.val == k) {
                queue.add(node);
                seen.add(node);
            }
        }

        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node != null) {
                if (graph.get(node).size() <= 1)
                    return node.val;
                for (TreeNode nei: graph.get(node)) {
                    if (!seen.contains(nei)) {
                        seen.add(nei);
                        queue.add(nei);
                    }
                }
            }
        }
        throw null;
    }

    public void dfs(Map<TreeNode, List<TreeNode>> graph, TreeNode node, TreeNode parent) {
        if (node != null) {
            if (!graph.containsKey(node)) graph.put(node, new LinkedList<TreeNode>());
            if (!graph.containsKey(parent)) graph.put(parent, new LinkedList<TreeNode>());
            graph.get(node).add(parent);
            graph.get(parent).add(node);
            dfs(graph, node.left, node);
            dfs(graph, node.right, node);
        }
    }
}