742. 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.

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.

Analysis:
这套题还挺有意义的，拓宽了做题思路。
Use DFS to convert binary tree to undirected graph, then use BFS to get the closest leaf.

Solution:
Check leaf node first before getting its neighbors. If there is only one node, get node from null would throw error.

class Solution {
    TreeNode startNode;
    public int findClosestLeaf(TreeNode root, int k) {
        Map<TreeNode, List<TreeNode>> graph = new HashMap<>();
        Queue<TreeNode> queue = new LinkedList<>();
        buildGraph(root, null, graph, k);
        queue.offer(startNode);
        Set<TreeNode> seen = new HashSet<>();
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur.left == null && cur.right == null)
                    return cur.val;
            for (TreeNode node: graph.get(cur)) {
                if (seen.add(node)) {
                    queue.offer(node);    
                }
                
            }
        }

        throw null;
    }

    private void buildGraph(TreeNode node, TreeNode parent, Map<TreeNode, List<TreeNode>> graph, int k) {
        if (node == null) return;
        if (parent != null) {
            graph.computeIfAbsent(parent, n -> new ArrayList<>()).add(node);
            graph.computeIfAbsent(node, n -> new ArrayList<>()).add(parent);
        }
        if (node.val == k) 
            startNode = node;
        buildGraph(node.left, node, graph, k);
        buildGraph(node.right, node, graph, k);
    }
}