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 voyagewe 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
给定一个有
N 个节点的二叉树，每个节点都有一个不同于其他节点且处于 {1, ..., N} 中的值。
通过交换节点的左子节点和右子节点，可以翻转该二叉树中的节点。
考虑从根节点开始的先序遍历报告的 N 值序列。将这一 N 值序列称为树的行程。
（回想一下，节点的先序遍历意味着我们报告当前节点的值，然后先序遍历左子节点，再先序遍历右子节点。）
我们的目标是翻转最少的树中节点，以便树的行程与给定的行程 voyage 相匹配。
如果可以，则返回翻转的所有节点的值的列表。你可以按任何顺序返回答案。
如果不能，则返回列表 [-1]。
示例 1：
输入：root = [1,2], voyage = [2,1] 输出：[-1]
示例 2：
输入：root = [1,2,3], voyage = [1,3,2] 输出：[1]
示例 3：
输入：root = [1,2,3], voyage = [1,2,3] 输出：[]
提示：
1 <= N <= 100
20ms
1 /**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 * public var val: Int
5 * public var left: TreeNode?
6 * public var right: TreeNode?
7 * public init(_ val: Int) {
8 * self.val = val
9 * self.left = nil
10 * self.right = nil
11 * }
12 * }
13 */
14 class Solution {
15 var p:Int = 0
16 var fed:[Int] = [Int]()
17 func flipMatchVoyage(_ root: TreeNode?, _ voyage: [Int]) -> [Int] {
18 p = 0
19 if dfs(root, voyage)
20 {
21 return fed
22 }
23 else
24 {
25 fed = [Int]()
26 fed.append(-1);
27 return fed;
28 }
29 }
30
31 func dfs(_ cur: TreeNode?, _ voyage: [Int]) -> Bool
32 {
33 if cur == nil {return true}
34 if voyage[p] != cur!.val {return false}
35 p += 1
36 if cur!.left == nil
37 {
38 return dfs(cur?.right, voyage)
39 }
40 if cur!.right == nil
41 {
42 return dfs(cur?.left, voyage)
43 }
44
45 if voyage[p] == cur!.left!.val
46 {
47 var res:Bool = dfs(cur?.left, voyage)
48 if !res
49 {
50 return false
51 }
52 return dfs(cur?.right, voyage)
53 }
54 else
55 {
56 fed.append(cur!.val)
57 var res:Bool = dfs(cur?.right, voyage)
58 if !res
59 {
60 return false
61 }
62 return dfs(cur?.left, voyage)
63 }
64 }
65 }
原文地址：https://www.cnblogs.com/strengthen/p/10228410.html