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

Solution 1: Lowest Common Ancestor + DFS

Thinking

The unique path goes through the LCA. Upward edges become U; the descent uses L/R. Three tree walks are fine for \(n \le 10^5\).

Find the LCA, DFS both directions from it, replace the start path by Us, and concatenate.

We can first find the lowest common ancestor of nodes \(\textit{startValue}\) and \(\textit{destValue}\), denoted as \(\textit{node}\). Then, starting from \(\textit{node}\), we find the paths to \(\textit{startValue}\) and \(\textit{destValue}\) respectively. The path from \(\textit{startValue}\) to \(\textit{node}\) will consist of a number of \(\textit{U}\)s, and the path from \(\textit{node}\) to \(\textit{destValue}\) will be the \(\textit{path}\). Finally, we concatenate these two paths.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptJavaScript

# 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:
        def lca(node: Optional[TreeNode], p: int, q: int):
            if node is None or node.val in (p, q):
                return node
            left = lca(node.left, p, q)
            right = lca(node.right, p, q)
            if left and right:
                return node
            return left or right

        def dfs(node: Optional[TreeNode], x: int, path: List[str]):
            if node is None:
                return False
            if node.val == x:
                return True
            path.append("L")
            if dfs(node.left, x, path):
                return True
            path[-1] = "R"
            if dfs(node.right, x, path):
                return True
            path.pop()
            return False

        node = lca(root, startValue, destValue)

        path_to_start = []
        path_to_dest = []

        dfs(node, startValue, path_to_start)
        dfs(node, destValue, path_to_dest)

        return "U" * len(path_to_start) + "".join(path_to_dest)

class Solution {
    public String getDirections(TreeNode root, int startValue, int destValue) {
        TreeNode node = lca(root, startValue, destValue);
        StringBuilder pathToStart = new StringBuilder();
        StringBuilder pathToDest = new StringBuilder();
        dfs(node, startValue, pathToStart);
        dfs(node, destValue, pathToDest);
        return "U".repeat(pathToStart.length()) + pathToDest.toString();
    }

    private TreeNode lca(TreeNode node, int p, int q) {
        if (node == null || node.val == p || node.val == q) {
            return node;
        }
        TreeNode left = lca(node.left, p, q);
        TreeNode right = lca(node.right, p, q);
        if (left != null && right != null) {
            return node;
        }
        return left != null ? left : right;
    }

    private boolean dfs(TreeNode node, int x, StringBuilder path) {
        if (node == null) {
            return false;
        }
        if (node.val == x) {
            return true;
        }
        path.append('L');
        if (dfs(node.left, x, path)) {
            return true;
        }
        path.setCharAt(path.length() - 1, 'R');
        if (dfs(node.right, x, path)) {
            return true;
        }
        path.deleteCharAt(path.length() - 1);
        return false;
    }
}

/**
 * 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:
    string getDirections(TreeNode* root, int startValue, int destValue) {
        TreeNode* node = lca(root, startValue, destValue);
        string pathToStart, pathToDest;
        dfs(node, startValue, pathToStart);
        dfs(node, destValue, pathToDest);
        return string(pathToStart.size(), 'U') + pathToDest;
    }

private:
    TreeNode* lca(TreeNode* node, int p, int q) {
        if (node == nullptr || node->val == p || node->val == q) {
            return node;
        }
        TreeNode* left = lca(node->left, p, q);
        TreeNode* right = lca(node->right, p, q);
        if (left != nullptr && right != nullptr) {
            return node;
        }
        return left != nullptr ? left : right;
    }

    bool dfs(TreeNode* node, int x, string& path) {
        if (node == nullptr) {
            return false;
        }
        if (node->val == x) {
            return true;
        }
        path.push_back('L');
        if (dfs(node->left, x, path)) {
            return true;
        }
        path.back() = 'R';
        if (dfs(node->right, x, path)) {
            return true;
        }
        path.pop_back();
        return false;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func getDirections(root *TreeNode, startValue int, destValue int) string {
    var lca func(node *TreeNode, p, q int) *TreeNode
    lca = func(node *TreeNode, p, q int) *TreeNode {
        if node == nil || node.Val == p || node.Val == q {
            return node
        }
        left := lca(node.Left, p, q)
        right := lca(node.Right, p, q)
        if left != nil && right != nil {
            return node
        }
        if left != nil {
            return left
        }
        return right
    }
    var dfs func(node *TreeNode, x int, path *[]byte) bool
    dfs = func(node *TreeNode, x int, path *[]byte) bool {
        if node == nil {
            return false
        }
        if node.Val == x {
            return true
        }
        *path = append(*path, 'L')
        if dfs(node.Left, x, path) {
            return true
        }
        (*path)[len(*path)-1] = 'R'
        if dfs(node.Right, x, path) {
            return true
        }
        *path = (*path)[:len(*path)-1]
        return false
    }

    node := lca(root, startValue, destValue)
    pathToStart := []byte{}
    pathToDest := []byte{}
    dfs(node, startValue, &pathToStart)
    dfs(node, destValue, &pathToDest)
    return string(bytes.Repeat([]byte{'U'}, len(pathToStart))) + string(pathToDest)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function getDirections(root: TreeNode | null, startValue: number, destValue: number): string {
    const lca = (node: TreeNode | null, p: number, q: number): TreeNode | null => {
        if (node === null || [p, q].includes(node.val)) {
            return node;
        }
        const left = lca(node.left, p, q);
        const right = lca(node.right, p, q);

        return left && right ? node : (left ?? right);
    };

    const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
        if (node === null) {
            return false;
        }
        if (node.val === x) {
            return true;
        }
        path.push('L');
        if (dfs(node.left, x, path)) {
            return true;
        }
        path[path.length - 1] = 'R';
        if (dfs(node.right, x, path)) {
            return true;
        }
        path.pop();
        return false;
    };

    const node = lca(root, startValue, destValue);
    const pathToStart: string[] = [];
    const pathToDest: string[] = [];
    dfs(node, startValue, pathToStart);
    dfs(node, destValue, pathToDest);
    return 'U'.repeat(pathToStart.length) + pathToDest.join('');
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} startValue
 * @param {number} destValue
 * @return {string}
 */
var getDirections = function (root, startValue, destValue) {
    const lca = (node, p, q) => {
        if (node === null || [p, q].includes(node.val)) {
            return node;
        }
        const left = lca(node.left, p, q);
        const right = lca(node.right, p, q);

        return left && right ? node : (left ?? right);
    };

    const dfs = (node, x, path) => {
        if (node === null) {
            return false;
        }
        if (node.val === x) {
            return true;
        }
        path.push('L');
        if (dfs(node.left, x, path)) {
            return true;
        }
        path[path.length - 1] = 'R';
        if (dfs(node.right, x, path)) {
            return true;
        }
        path.pop();
        return false;
    };

    const node = lca(root, startValue, destValue);
    const pathToStart = [];
    const pathToDest = [];
    dfs(node, startValue, pathToStart);
    dfs(node, destValue, pathToDest);
    return 'U'.repeat(pathToStart.length) + pathToDest.join('');
};

Solution 2: Lowest Common Ancestor + DFS (Optimized)

Thinking

Solution 1 finds the LCA then walks twice more. Paths from the root share a prefix; stripping it is exactly “up to the LCA then down.” The extra LCA search disappears.

Two DFS strings, skip the common prefix of length \(i\), emit \((|start|-i)\) Us plus the destination suffix.

We can start from \(\textit{root}\), find the paths to \(\textit{startValue}\) and \(\textit{destValue}\), denoted as \(\textit{pathToStart}\) and \(\textit{pathToDest}\), respectively. Then, remove the longest common prefix of \(\textit{pathToStart}\) and \(\textit{pathToDest}\). At this point, the length of \(\textit{pathToStart}\) is the number of \(\textit{U}\)s in the answer, and the path of \(\textit{pathToDest}\) is the path in the answer. We just need to concatenate these two paths.