# 971. Flip Binary Tree To Match Preorder Traversal 翻转二叉树以匹配先序遍历

@TOC

## # 题目描述

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: []
``````

Note:

1. 1 <= N <= 100

## # 解题方法

### # 前序遍历

C++代码如下：

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int i = 0;
vector<int> res;
vector<int> flipMatchVoyage(TreeNode* root, vector<int>& voyage) {
res.clear();
if (dfs(root, voyage)) {
return res;
}
return {-1};
}
// can we get v by flip root?
bool dfs(TreeNode* root, vector<int>& v) {
if (!root) return true;
if (root->val != v[i++]) return false;
if (root->left && root->left->val == v[i]) {
return dfs(root->left, v) && dfs(root->right, v);
} else if (root->right && root->right->val == v[i]) {
if (root->left)
res.push_back(root->val);
return dfs(root->right, v) && dfs(root->left, v);
}
return !root->left && !root->right;
}
};
``````

## # 日期

2019 年 1 月 6 日 —— 打球打的腰酸背痛