Problem description

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.

(Recall that a preorder traversal of a node means we report the current node’s value, then preorder-traverse the left child, then preorder-traverse the right child.)

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: []

Solution

1. We can use DFS to solve the problem. We keep two attributes (nodes flipped and position in voyage) while doing depth first search. If the node is None, we return []. If it is not None and node.val != voyage[position], there is no way to achieve the match by flipping, we return [-1]. If node.val == voyage[position], we check whether the node.left is None or node.left.val == voyage[position + 1]. If left is none or the value does not equal to voyage[position + 1], we need to flip the node.val (add it to flipped). Then we recursively call right child node first, and then left child node (since node is flipped). Otherwise, we recursively call left child node first, and then right child node.

Below is my python implementation

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def flipMatchVoyage(self, root: TreeNode, voyage: List[int]) -> List[int]:
        self.flipped = []
        self.i = 0
        
        def dfs(root):
            if not root:
                return
            if root.val != voyage[self.i]:
                self.flipped = [-1]
                return
            self.i += 1
            if root.left and root.left.val != voyage[self.i]:
                self.flipped.append(root.val)
                dfs(root.right)
                dfs(root.left)
            else:
                dfs(root.left)
                dfs(root.right)
        
        dfs(root)
        if self.flipped and self.flipped[0] == -1:
            return self.flipped
        else:
            return self.flipped