[Lintcode] Problem 854 - Closest Leaf in a Binary Tree

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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. If there are multiple cases, you should follow these priorities:

  1. The leaf node is in the left subtree of the node with k;
  2. The leaf node is in the right subtree of the node with k;
  3. The leaf node is not in the subtree of the node with k.

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 a node in the given binary tree for which node.val == k.

Example

No.1

Input: {1, 3, 2}, k = 1

Output: 3

Explanation:

1
2
3
  1
 / \
3   2

No.2

Input: {1}, k = 1

Output: 1

Code

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public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode() {}
    TreeNode(int val) { this.val = val; }
    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}
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private TreeNode start;
private Map<TreeNode, TreeNode> connection = new HashMap<>();

public int findClosestLeaf(TreeNode root, int k) {
    Set<TreeNode> visit = new HashSet<>();
    Queue<TreeNode> queue = new LinkedList<>();

    dfs(null, root, k);
    queue.offer(start);
    visit.add(start);

    while (!queue.isEmpty()) {
        TreeNode node = queue.poll();

        if (node.left == null && node.right == null)
            return node.val;

        if (node.left != null && !visit.contains(node.left)) {
            queue.offer(node.left);
            visit.add(node.left);
        }

        if (node.right != null && !visit.contains(node.right)) {
            queue.offer(node.right);
            visit.add(node.right);
        }

        if (connection.containsKey(node) && !visit.contains(connection.get(node))) {
            queue.offer(connection.get(node));
            visit.add(connection.get(node));
        }
    }

    return 0;
}

private void dfs(TreeNode parent, TreeNode node, int k) {
    if (node == null)
        return;

    if (node.val == k)
        start = node;
    
    if (parent != null) {
        connection.put(node, parent);
    }

    dfs(node, node.left, k);
    dfs(node, node.right, k);
}