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 <= 1051 <= Node.val <= n- All the values in the tree are unique.
1 <= startValue, destValue <= nstartValue != destValue
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
| # 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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
| /**
* 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);
}
}
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
| /**
* 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();
}
}
}
};
|
