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2096. Step-By-Step Directions From a Binary Tree Node to Another

Description

You are given the root of a binary tree with n nodes. Each node is uniquely assigned a value from 1 to n. You are also given an integer startValue representing the value of the start node s, and a different integer destValue representing the value of the destination node t.

Find the shortest path starting from node s and ending at node t. Generate step-by-step directions of such path as a string consisting of only the uppercase letters 'L', 'R', and 'U'. Each letter indicates a specific direction:

  • 'L' means to go from a node to its left child node.
  • 'R' means to go from a node to its right child node.
  • 'U' means to go from a node to its parent node.

Return the step-by-step directions of the shortest path from node s to node t.

 

Example 1:

Input: root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6
Output: "UURL"
Explanation: The shortest path is: 3 → 1 → 5 → 2 → 6.

Example 2:

Input: root = [2,1], startValue = 2, destValue = 1
Output: "L"
Explanation: The shortest path is: 2 → 1.

 

Constraints:

  • The number of nodes in the tree is n.
  • 2 <= n <= 105
  • 1 <= Node.val <= n
  • All the values in the tree are unique.
  • 1 <= startValue, destValue <= n
  • startValue != destValue

Solutions

  • /**
 * Definition for a binary tree node.
 * 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;
 *     }
 * }
 */
class Solution {
    private Map<Integer, List<List<String>>> edges;
    private Set<Integer> visited;
    private String ans;

    public String getDirections(TreeNode root, int startValue, int destValue) {
        edges = new HashMap<>();
        visited = new HashSet<>();
        ans = null;
        traverse(root);
        dfs(startValue, destValue, new ArrayList<>());
        return ans;
    }

    private void traverse(TreeNode root) {
        if (root == null) {
            return;
        }
        if (root.left != null) {
            edges.computeIfAbsent(root.val, k -> new ArrayList<>())
                .add(Arrays.asList(String.valueOf(root.left.val), "L"));
            edges.computeIfAbsent(root.left.val, k -> new ArrayList<>())
                .add(Arrays.asList(String.valueOf(root.val), "U"));
        }
        if (root.right != null) {
            edges.computeIfAbsent(root.val, k -> new ArrayList<>())
                .add(Arrays.asList(String.valueOf(root.right.val), "R"));
            edges.computeIfAbsent(root.right.val, k -> new ArrayList<>())
                .add(Arrays.asList(String.valueOf(root.val), "U"));
        }
        traverse(root.left);
        traverse(root.right);
    }

    private void dfs(int start, int dest, List<String> t) {
        if (visited.contains(start)) {
            return;
        }
        if (start == dest) {
            if (ans == null || ans.length() > t.size()) {
                ans = String.join("", t);
            }
            return;
        }
        visited.add(start);
        if (edges.containsKey(start)) {
            for (List<String> item : edges.get(start)) {
                t.add(item.get(1));
                dfs(Integer.parseInt(item.get(0)), dest, t);
                t.remove(t.size() - 1);
            }
        }
    }
}

  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    unordered_map<int, vector<pair<int, char>>> edges;
    unordered_set<int> visited;
    string ans;

    string getDirections(TreeNode* root, int startValue, int destValue) {
        ans = "";
        traverse(root);
        string t = "";
        dfs(startValue, destValue, t);
        return ans;
    }

    void traverse(TreeNode* root) {
        if (!root) return;
        if (root->left) {
            edges[root->val].push_back({root->left->val, 'L'});
            edges[root->left->val].push_back({root->val, 'U'});
        }
        if (root->right) {
            edges[root->val].push_back({root->right->val, 'R'});
            edges[root->right->val].push_back({root->val, 'U'});
        }
        traverse(root->left);
        traverse(root->right);
    }

    void dfs(int start, int dest, string& t) {
        if (visited.count(start)) return;
        if (start == dest) {
            if (ans == "" || ans.size() > t.size()) ans = t;
            return;
        }
        visited.insert(start);
        if (edges.count(start)) {
            for (auto& item : edges[start]) {
                t += item.second;
                dfs(item.first, dest, t);
                t.pop_back();
            }
        }
    }
};

  • # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def getDirections(
        self, root: Optional[TreeNode], startValue: int, destValue: int
    ) -> str:
        edges = defaultdict(list)
        ans = None
        visited = set()

        def traverse(root):
            if not root:
                return
            if root.left:
                edges[root.val].append([root.left.val, 'L'])
                edges[root.left.val].append([root.val, 'U'])
            if root.right:
                edges[root.val].append([root.right.val, 'R'])
                edges[root.right.val].append([root.val, 'U'])
            traverse(root.left)
            traverse(root.right)

        def dfs(start, dest, t):
            nonlocal ans
            if start in visited:
                return
            if start == dest:
                if ans is None or len(ans) > len(t):
                    ans = ''.join(t)
                return
            visited.add(start)
            for d, k in edges[start]:
                t.append(k)
                dfs(d, dest, t)
                t.pop()

        traverse(root)
        dfs(startValue, destValue, [])
        return ans

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