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This approach involves finding the path from the start node to the root and the destination node to the root. Using these paths, the Lowest Common Ancestor (LCA) of the start and destination nodes is determined. This allows us to break down the path from start to destination as moving up towards the root to the LCA, then down towards the destination. Once these steps are identified, we can replace moves with correct directions ('L', 'R', 'U').
Time Complexity: O(n), where n is the number of nodes, as each node could be traversed in the worst-case.
Space Complexity: O(h), where h is the height of the tree due to recursion stack and storing paths.
1import java.util.*;
2
3class TreeNode {
4 int val;
5 TreeNode left;
6 TreeNode right;
7
8 TreeNode(int x) {
9 val = x;
10 left = null;
11 right = null;
12 }
13}
14
15public class Solution {
16 private boolean pathToNode(TreeNode root, int value, List<Character> path) {
17 if (root == null) return false;
18 if (root.val == value) return true;
19
20 path.add('L');
21 if (pathToNode(root.left, value, path)) return true;
22 path.remove(path.size() - 1);
23
24 path.add('R');
25 if (pathToNode(root.right, value, path)) return true;
26 path.remove(path.size() - 1);
27
28 return false;
29 }
30
31 public String getDirections(TreeNode root, int startValue, int destValue) {
32 List<Character> startPath = new ArrayList<>();
33 List<Character> destPath = new ArrayList<>();
34
35 pathToNode(root, startValue, startPath);
36 pathToNode(root, destValue, destPath);
37
38 int i = 0;
39 while (i < startPath.size() && i < destPath.size() && startPath.get(i) == destPath.get(i)) i++;
40 StringBuilder sb = new StringBuilder();
41
42 for (int j = i; j < startPath.size(); j++) {
43 sb.append('U');
44 }
45 for (int j = i; j < destPath.size(); j++) {
46 sb.append(destPath.get(j));
47 }
48
49 return sb.toString();
50 }
51
52 public static void main(String[] args) {
53 TreeNode root = new TreeNode(5);
54 root.left = new TreeNode(1);
55 root.right = new TreeNode(2);
56 root.left.left = new TreeNode(3);
57 root.right.left = new TreeNode(6);
58 root.right.right = new TreeNode(4);
59
60 Solution sol = new Solution();
61 System.out.println(sol.getDirections(root, 3, 6)); // Output: "UURL"
62 }
63}The Java solution mirrors the logic of the Python and C++ implementations. It uses a recursive helper method to determine paths to the specified nodes and then constructs the necessary direction steps. The solution accounts for shared path segments leading to the lowest common ancestor.
This approach uses a DFS to build a parent map of nodes first. With this map, we can easily track the movement from any node to any other node. We then extract paths from startValue and destValue up to a common ancestor. The directions are computed through knowledge of parent-child relationships and mapped accordingly using 'L', 'R', and 'U'.
Time Complexity: O(n), traversing all nodes for setup. Space Complexity: O(n) due to the parent mapping.
1using System;
2using System.Collections.Generic;
3
public class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x) { val = x; }
}
public class Solution {
private Dictionary<int, TreeNode> parentMap = new Dictionary<int, TreeNode>();
private void DFS(TreeNode node, TreeNode parent) {
if (node == null) return;
parentMap[node.val] = parent;
DFS(node.left, node);
DFS(node.right, node);
}
private List<char> GetPath(int value) {
List<char> path = new List<char>();
while (parentMap.ContainsKey(value) && parentMap[value] != null) {
TreeNode parent = parentMap[value];
if (parent.left != null && parent.left.val == value) {
path.Add('L');
} else if (parent.right != null && parent.right.val == value) {
path.Add('R');
}
value = parent.val;
}
path.Reverse();
return path;
}
public string GetDirections(TreeNode root, int startValue, int destValue) {
DFS(root, null);
List<char> startPath = GetPath(startValue);
List<char> destPath = GetPath(destValue);
int i = 0;
while (i < startPath.Count && i < destPath.Count && startPath[i] == destPath[i]) {
i++;
}
return new string('U', startPath.Count - i) + string.Concat(destPath.GetRange(i, destPath.Count - i));
}
public static void Main(string[] args) {
TreeNode root = new TreeNode(5);
root.left = new TreeNode(1);
root.right = new TreeNode(2);
root.left.left = new TreeNode(3);
root.right.left = new TreeNode(6);
root.right.right = new TreeNode(4);
Solution sol = new Solution();
Console.WriteLine(sol.GetDirections(root, 3, 6)); // Output: "UURL"
}
}This C# implementation employs DFS to build a dictionary representing parent relationships, similar to the Python solution. It calculates paths from start and destination nodes up to their most recent common ancestor and uses these details to derive instructions for the needed path using 'L', 'R', and 'U'.