LeetCode-863. All Nodes Distance K in Binary Tree

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​​.

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.

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2Output: [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​​.

题解：

dfs构造map，bfs找值

class Solution {public: static void bfs (vector &visit, vector &ans, int t, int d, vector> &map, int k) { if (d == k - 1) { for (int i = 0; i < map[t].size(); i++) { if (visit[map[t][i]] == false) { ans.push_back(map[t][i]); } } } if (d != k - 1) { for (int i = 0; i < map[t].size(); i++) { if (visit[map[t][i]] == false) { visit[map[t][i]] = true; bfs(visit, ans, map[t][i], d + 1, map, k); } } } } vector distanceK(TreeNode* root, TreeNode* target, int K) { vector ans; vector visit(501, false); if (root == NULL) { return ans; } if (K == 0) { ans.push_back(target->val); return ans; } vector> map(501, ans); stack dfs; dfs.push(root); while (dfs.empty() == false) { TreeNode* t = dfs.top(); dfs.pop(); if (t->left != NULL) { dfs.push(t->left); map[t->val].push_back(t->left->val); map[t->left->val].push_back(t->val); } if (t->right != NULL) { dfs.push(t->right); map[t->val].push_back(t->right->val); map[t->right->val].push_back(t->val); } } visit[target->val] = true; bfs(visit, ans, target->val, 0, map, K); return ans; }};

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