Leetcode 971. Flip Binary Tree To Match Preorder Traversal (二叉树遍历好题)

971. Flip Binary Tree To Match Preorder Traversal
     Medium

You are given the root of a binary tree with n nodes, where each node is uniquely assigned a value from 1 to n. You are also given a sequence of n values voyage, which is the desired pre-order traversal of the binary tree.

Any node in the binary tree can be flipped by swapping its left and right subtrees. For example, flipping node 1 will have the following effect:

Flip the smallest number of nodes so that the pre-order traversal of the tree matches voyage.

Return a list of the values of all flipped nodes. You may return the answer in any order. If it is impossible to flip the nodes in the tree to make the pre-order traversal match voyage, return the list [-1].

Example 1:

Input: root = [1,2], voyage = [2,1]
Output: [-1]
Explanation: It is impossible to flip the nodes such that the pre-order traversal matches voyage.
Example 2:

Input: root = [1,2,3], voyage = [1,3,2]
Output: [1]
Explanation: Flipping node 1 swaps nodes 2 and 3, so the pre-order traversal matches voyage.
Example 3:

Input: root = [1,2,3], voyage = [1,2,3]
Output: []
Explanation: The tree’s pre-order traversal already matches voyage, so no nodes need to be flipped.

Constraints:

The number of nodes in the tree is n.
n == voyage.length
1 <= n <= 100
1 <= Node.val, voyage[i] <= n
All the values in the tree are unique.
All the values in voyage are unique.

解法1：
参考的labuladong的思路: 用 traverse 函数遍历整棵二叉树，对比前序遍历结果，如果节点的值对不上，就无解；如果子树对不上 voyage，就尝试翻转子树。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> flipMatchVoyage(TreeNode* root, vector<int>& voyage) {
        int index = 0;
        helper(root, voyage, index);
        if (isValid) return res;
        return {-1};
    }
private:
    vector<int> res;
    bool isValid = true;
    void helper(TreeNode *root, vector<int>& voyage, int &index) {
        if (!root) return;
        if (root->val != voyage[index]) {
            isValid = false;
        }
        index++;
        if (root->left && root->left->val != voyage[index]) {
            res.push_back(root->val);
            swap(root->left, root->right);            
        }
        helper(root->left, voyage, index);
        helper(root->right, voyage, index);
    }
};