863. all nodes distance k in binary tree

Use HashMap,先用bfs或dfs将树变成一个图,之后根据target node的neighbors进行逐层搜索,直到distance == K。
通过HashSet avoid visited nodes。

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* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class {
Map<TreeNode, HashSet<TreeNode>> map = new HashMap<>();
public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
List<Integer> list = new ArrayList<>();
if (root == null) return list;
buildMap(root);
Queue<TreeNode> q = new LinkedList<>();
q.offer(target);
Set<TreeNode> visited = new HashSet<>();
visited.add(target);
boolean found = false;
int dis = 0;
while (!q.isEmpty()) {
int len = q.size();
for (int size = 0; size < len; size++) {
TreeNode cur = q.poll();
if (dis == K) {
list.add(cur.val);
found = true;
}else {
if (!map.get(cur).isEmpty()) {
for (TreeNode t : map.get(cur)) {
if (!visited.contains(t)) {
q.offer(t);
visited.add(t);
}
}
}
}
}
if (found) return list;
dis++;
}
return list;
}
public void buildMap(TreeNode node) {
Queue<TreeNode> q = new LinkedList<>();
q.offer(node);
map.put(node, new HashSet<TreeNode>());
while (!q.isEmpty()) {
TreeNode cur = q.poll();
if (cur.left != null) {
map.put(cur.left, new HashSet<TreeNode>());
map.get(cur).add(cur.left);
map.get(cur.left).add(cur);
q.offer(cur.left);
}
if (cur.right != null) {
map.put(cur.right, new HashSet<TreeNode>());
map.get(cur).add(cur.right);
map.get(cur.right).add(cur);
q.offer(cur.right);
}
}
}
}