[leetcode]971. Flip Binary Tree To Match Preorder Traversal

Analysis

昨天和今天天气都很好鸭~嘻嘻—— [每天刷题并不难0.0]

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].

Explanation:

先序遍历给定的树，同时遍历voyage，如果节点值等于voyage里的值，但是左儿子的值不等于voyage里的值，则调换左右儿子节点，继续遍历

Implement

/**
 * 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:
    vector<int> flipMatchVoyage(TreeNode* root, vector<int>& voyage) {
        return dfs(root, voyage)?res:vector<int>{-1};
    }
    bool dfs(TreeNode* node, vector<int>& voyage){
        if(!node)
            return true;
        if(node->val != voyage[i++])
            return false;
        if(node->left && node->left->val != voyage[i]){
            res.push_back(node->val);
            return dfs(node->right, voyage) && dfs(node->left, voyage);
        }
        return dfs(node->left, voyage) && dfs(node->right, voyage);
    }
private:
    vector<int> res;
    int i;
};