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authorjack <[email protected]>2024-06-14 13:13:15 -0600
committerjack <[email protected]>2024-06-14 13:13:15 -0600
commitd343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch)
tree96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /all_nodes_binary_tree.cpp
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+// 863. All Nodes Distance K in Binary Tree
+#include "leetcode.h"
+
+/*
+ * given the 'root' of a binary tree, the value of a target node 'target', and
+ * an integer 'k', return an array of the values of all nodes that have a
+ * distance 'k' from the target node. you can return the answer in any order.
+ */
+
+class Solution {
+ void dfs(TreeNode *root, unordered_map<TreeNode *, TreeNode *> &back_edge,
+ TreeNode *target) {
+ if (!root || root == target)
+ return;
+ if (root->left) {
+ back_edge[root->left] = root;
+ dfs(root->left, back_edge, target);
+ }
+ if (root->right) {
+ back_edge[root->right] = root;
+ dfs(root->right, back_edge, target);
+ }
+ }
+
+public:
+ vector<int> distanceK(TreeNode *root, TreeNode *target, int k) {
+ unordered_map<TreeNode *, TreeNode *> back_edge;
+ unordered_set<TreeNode *> visited;
+ vector<int> ans;
+ dfs(root, back_edge, target);
+ queue<TreeNode *> q;
+ q.push(target);
+ while (!q.empty() && k >= 0) {
+ int s = q.size();
+ while (s--) {
+ TreeNode *curr = q.front();
+ int val = curr->val;
+ q.pop();
+ visited.insert(curr);
+ if (!k)
+ ans.push_back(val);
+ if (!visited.count(back_edge[curr]) && back_edge[curr])
+ q.push(back_edge[curr]);
+ if (!visited.count(curr->left) && curr->left)
+ q.push(curr->left);
+ if (!visited.count(curr->right) && curr->right)
+ q.push(curr->right);
+ }
+ k--;
+ }
+ return ans;
+ }
+};