Flip Binary Tree To Match Preorder Traversal

LeetCode Q 971 - Flip Binary Tree To Match Preorder Traversal

Given a binary tree with N nodes, each node has a different value from {1, ..., N}.
A node in this binary tree can be flipped by swapping the left child and the right child of that node.
Consider the sequence of N values reported by a preorder traversal starting from the root. Call such a sequence of N values the voyage of the tree.
Our goal is to flip the least number of nodes in the tree so that the voyage of the tree matches the voyage we are given.
If we can do so, then return a list of the values of all nodes flipped. You may return the answer in any order.
If we cannot do so, then return the list [-1].

Example 1: Input: root = [1,2], voyage = [2,1] ; Output: [-1]
Example 2: Input: root = [1,2,3], voyage = [1,3,2] ; Output: [1]
Example 3: Input: root = [1,2,3], voyage = [1,2,3] ; Output: []

Note: 1 <= N <= 100

Solution: DFS

Code:

List<Integer> res;
int index = 0;
public List<Integer> flipMatchVoyage(TreeNode root, int[] voyage) {
  res = new ArrayList<>();
  return dfs(root) ? res : Arrays.asList(-1);
}

private boolean dfs(TreeNode root, int[] voyage) {
  if (root == null) return true;

  if (voyage[index] != root.val) return false;

  index++;

  if (root.left != null && root.left.val != voyage[index]) {
    res.add(root); 
    return dfs(root.right, voyage) && dfs(root.left, voyage); 
  }

  return dfs(root.left, voyage) && dfs(root.right, voyage);
}