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Problem solution · Java

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

Step-By-Step Directions From a Binary Tree Node to Another: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
30 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Step-By-Step Directions From a Binary Tree Node to Another, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 30 lines of Java from the credited upstream file 2096-2.java.
  • The implementation keeps its working state in language-native values and containers.
  • 1 loop block detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeStep-By-Step Directions From a Binary Tree Node to Another · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public String getDirections(TreeNode root, int startValue, int destValue) {    StringBuilder pathToStart = new StringBuilder();    StringBuilder pathToDest = new StringBuilder();     dfs(root, startValue, pathToStart);    dfs(root, destValue, pathToDest);     while (pathToStart.length() > 0 && pathToDest.length() > 0 &&           pathToStart.charAt(pathToStart.length() - 1) ==               pathToDest.charAt(pathToDest.length() - 1)) {      pathToStart.setLength(pathToStart.length() - 1);      pathToDest.setLength(pathToDest.length() - 1);    }     return "U".repeat(pathToStart.length()) + pathToDest.reverse().toString();  }   // Builds the string in reverse order to avoid creating a new copy.  private boolean dfs(TreeNode root, int val, StringBuilder sb) {    if (root.val == val)      return true;    if (root.left != null && dfs(root.left, val, sb))      sb.append("L");    else if (root.right != null && dfs(root.right, val, sb))      sb.append("R");    return sb.length() > 0;  }}

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