leetcode 863. All Nodes Distance K in Binary Tree

We are given a binary tree (with root node root), a target node, and an integer value K.node

Return a list of the values of all nodes that have a distance K from the target node. The answer can be returned in any order.3d

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2
Output: [7,4,1]
Explanation: 
The nodes that are a distance 2 from the target node (with value 5)
have values 7, 4, and 1.

Note that the inputs "root" and "target" are actually TreeNodes.
The descriptions of the inputs above are just serializations of these objects.

Note:

The given tree is non-empty.
Each node in the tree has unique values 0 <= node.val <= 500.
The target node is a node in the tree.
0 <= K <= 1000.

思路：將二叉樹轉換爲無向圖，而後進行dfs記錄深度。注意，對無向圖進行遍歷的時候要記錄父節點，又不會有重複。code

/**
 * 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:
    map<int, vector<int>> edge;
    vector<int> ans;
    void dfs(TreeNode* root) {
        if (root == nullptr) return;
        if (root->left != nullptr) {
            dfs(root->left);
            edge[root->val].push_back(root->left->val);
            edge[root->left->val].push_back(root->val);
        }
        if (root->right != nullptr) {
            dfs(root->right); 
            edge[root->val].push_back(root->right->val);
            edge[root->right->val].push_back(root->val);
              
        }
    }
    void dfs2(int u, int fa, int dep, int K) {
        if (dep == K) {
            ans.push_back(u);
            return;
        }
        for (int i = 0; i < edge[u].size(); ++i) {
            int v = edge[u][i];
            if (v == fa) continue;
            dfs2(v, u, dep+1, K);
        }
    }
    vector<int> distanceK(TreeNode* root, TreeNode* target, int K) {
        dfs(root);
        dfs2(target->val, -1, 0, K);
        return ans;
    }
};