971. Flip Binary Tree To Match Preorder Traversal

T: O(n)
S: O(n)

/**
 * 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 {
    public List<Integer> flipMatchVoyage(TreeNode root, int[] voyage) {
        List<Integer> result = new ArrayList<>();
        int[] i = new int[1];
        boolean res = dfs(root, voyage, i, result);
        if (!res) {
            return List.of(-1);
        }
        return result;
    }
    private boolean dfs(TreeNode root, int[] voyage, int[] i, List<Integer> result) {
        if (root == null) {
            return true;
        }
        if (root.val != voyage[i[0]++]) { // root val is wrong, means can't flip to make it
            return false;
        }
        if (root.left != null && root.left.val != voyage[i[0]]) { // left not the same, do swap
            TreeNode temp = root.left;
            root.left = root.right;
            root.right = temp;
            result.add(root.val);
        }
        return dfs(root.left, voyage, i, result) && dfs(root.right, voyage, i, result);
    }
}

/***
T: O(n)
S: O(n)

 */