medium Path With Given Sequence
Problem Statement
Given a binary tree and a number sequence, find if the sequence is present as a root-to-leaf path in the given tree.
Constraints:
1 <= arr.length <= 50000 <= arr[i] <= 9- Each node's value is between
[0 - 9].
Try it yourself
Try solving this question here:
✅ Solution Path With Given Sequence
Problem Statement
Given a binary tree and a number sequence, find if the sequence is present as a root-to-leaf path in the given tree.
Constraints:
1 <= arr.length <= 50000 <= arr[i] <= 9- Each node's value is between
[0 - 9].
Solution
This problem follows the Binary Tree Path Sum pattern. We can follow the same DFS approach and additionally, track the element of the given sequence that we should match with the current node. Also, we can return false as soon as we find a mismatch between the sequence and the node value.
Here is the visual representation of the algorithm:
Code
Here is the code for this algorithm:
import java.util.*;
// class TreeNode {
// int val;
// TreeNode left;
// TreeNode right;
// TreeNode(int x) {
// val = x;
// }
// };
class Solution {
public boolean findPath(TreeNode root, int[] sequence) {
if (root == null) return sequence.length == 0;
return findPathRecursive(root, sequence, 0);
}
private static boolean findPathRecursive(
TreeNode currentNode,
int[] sequence,
int sequenceIndex
) {
if (currentNode == null) return false;
if (
sequenceIndex >= sequence.length ||
currentNode.val != sequence[sequenceIndex]
) return false;
// if the current node is a leaf, add it is the end of the sequence, we have found
// a path!
if (
currentNode.left == null &&
currentNode.right == null &&
sequenceIndex == sequence.length - 1
) return true;
// recursively call to traverse the left and right sub-tree
// return true if any of the two recursive call return true
return (
findPathRecursive(currentNode.left, sequence, sequenceIndex + 1) ||
findPathRecursive(currentNode.right, sequence, sequenceIndex + 1)
);
}
public static void main(String[] args) {
Solution sol = new Solution();
TreeNode root = new TreeNode(1);
root.left = new TreeNode(0);
root.right = new TreeNode(1);
root.left.left = new TreeNode(1);
root.right.left = new TreeNode(6);
root.right.right = new TreeNode(5);
System.out.println(
"Tree has path sequence: " + sol.findPath(root, new int[] { 1, 0, 7 })
);
System.out.println(
"Tree has path sequence: " + sol.findPath(root, new int[] { 1, 1, 6 })
);
}
}
Time Complexity
The time complexity of the above algorithm is O(N), where ‘N’ is the total number of nodes in the tree. This is due to the fact that we traverse each node once.
Space Complexity
The space complexity of the above algorithm will be O(N) in the worst case. This space will be used to store the recursion stack. The worst case will happen when the given tree is a linked list (i.e., every node has only one child).
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Progressively stronger hints — you still solve it.
I'm working on the problem **Path With Given Sequence** (DSA). Give me a HINT LADDER: start with the tiniest nudge, then wait. Only reveal the next, stronger hint when I ask. Do NOT show the full solution unless I type 'show solution'. Keep me doing the thinking. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.
See the technique, not just code.
Explain the optimal approach to **Path With Given Sequence** with a VISUAL walkthrough: trace it on a small concrete example using ASCII art / a step-by-step diagram, narrate what changes each step, then give time & space complexity with a one-line derivation. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.
Catch bugs, edge cases, sub-optimality.
I'll paste my solution to **Path With Given Sequence**. Review it for correctness, missed edge cases, and time/space complexity, then coach me toward the optimal — don't just rewrite it. Ask me to paste my code now. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.
Lock in recognition with look-alikes.
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