LC 971. Flip Binary Tree To Match Preorder Traversal

最新推荐文章于 2025-09-07 19:33:12 发布

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最新推荐文章于 2025-09-07 19:33:12 发布

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CC 4.0 BY-SA版权

文章标签： c/c++ runtime

原文链接：http://www.cnblogs.com/ethanhong/p/10301723.html

本文介绍了一种算法，旨在通过最少次数的节点翻转使二叉树的前序遍历与给定序列相匹配。文章提供了一个C++实现方案，该方案在在线提交中运行速度超过99.80%，并详细解释了递归辅助函数的逻辑，用于判断是否可以成功翻转，并记录翻转的节点。

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].

Runtime: 4 ms, faster than 99.80% of C++ online submissions for Flip Binary Tree To Match Preorder Traversal.

//
// Created by yuxi on 2019-01-18.
//

#include <vector>
#include <unordered_map>
using namespace std;


/**
 * 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 {
private:
  unordered_map<int,int> mp;
public:
  vector<int> flipMatchVoyage(TreeNode* root, vector<int>& voyage) {
    for(int i=0; i<voyage.size(); i++) mp[voyage[i]] = i;
    vector<int> ret;
    bool flag = helper(ret, root, 0);
    if(!flag) return {-1};
    return ret;
  }
  bool helper(vector<int>& ret, TreeNode* root, int idx){
    if(!root) return true;
    if(mp[root->val] != idx) return false;
    int left = root->left ? mp[root->left->val] : -1;
    int right = root->right ? mp[root->right->val] : -1;
    if((left >= 0 && left < mp[root->val]) || (right >= 0 && right < mp[root->val])) return false;
    if(left == -1 && right == -1) return true;
    else if(left == -1 && right != -1) {
      if(right != idx+1) return false;
      return helper(ret, root->right, mp[root->right->val]);
    }
    else if(left != -1 && right == -1) {
      if(left != idx+1) return false;
      return helper(ret, root->left, mp[root->left->val]);
    }
    else if(left > right) {
      if(right != idx+1) return false;
      ret.push_back(root->val);
      return helper(ret, root->right, idx+1) && helper(ret, root->left, mp[root->left->val]);
    }
    else {
      if(left != idx+1) return false;
      return helper(ret, root->left, idx+1) && helper(ret, root->right, mp[root->right->val]);
    }
    }
};