Leetcode 971. Flip Binary Tree To Match Preorder Traversal

Posted 2022-12-02 逆風的薔薇

题目

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

分析

以前序遍历二叉树，求能匹配对应序列时需交换左右子树的最少节点集。

考察深度优先搜索。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode 
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x)  val = x; 
 * 
 */
class Solution 
    public List<Integer> flipMatchVoyage(TreeNode root, int[] voyage) 
        List<Integer> flipPath = new ArrayList<>();
        int resultIdx = dfs(root, 0, flipPath, voyage);
        
        if (resultIdx == -1 || resultIdx < voyage.length-1) 
            return Arrays.asList(-1);
        
        return flipPath;
    

    public int dfs(TreeNode root, int idx, List<Integer> flipPath, int[] voyage) 
        if (root == null) 
            return idx;
        
        
        if (idx >= voyage.length) 
            return -1;
        

        if(root.val != voyage[idx]) 
            return -1;
        

        int left = dfs(root.left, idx + 1, flipPath, voyage);
        if (left != -1) 
            //Completed the leaf node in left child, then right child
            return dfs(root.right, left, flipPath, voyage);
        

        //Need to flip
        flipPath.add(root.val);

        int right = dfs(root.right, idx + 1, flipPath, voyage);
        if (right != -1) 
            return dfs(root.left, right, flipPath, voyage);
        

        return -1;