Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any one of them.
Two trees are duplicate if they have the same structure with same node values.
Example 1:
1
/ \
2 3
/ / \
4 2 4
/
4
The following are two duplicate subtrees:
2
/
4
and
4
Therefore, you need to return above trees' root in the form of a list.
代码
Approach #1 DFS
Time: O(N^2) where N is the number of nodes in the tree. We visit each node once, but each creation of serial may take O(N) work.
Space: O(N^2) the size of count
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
Map<String, Integer> count = new HashMap();
List<TreeNode> ans = new ArrayList();
collect(root, count, ans);
return ans;
}
private String collect(TreeNode node, Map<String, Integer> count, List<TreeNode> ans) {
if (node == null) return "#";
String serial = node.val + "," + collect(node.left, count, ans) + "," + collect(node.right, count, ans);
count.put(serial, count.getOrDefault(serial, 0) + 1);
if (count.get(serial) == 2) {
ans.add(node);
}
return serial;
}
}
Approach #2 Unique Identifier
Time: O(N log N)
Space: O(N)
class Solution {
int t;
Map<String, Integer> trees;
Map<Integer, Integer> count;
List<TreeNode> ans;
public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
t = 1;
trees = new HashMap();
count = new HashMap();
ans = new ArrayList();
lookup(root);
return ans;
}
private int loopup(TreeNode node) {
if (node == null) return 0;
String serial = node.val + "," + lookup(node.left) + "," + lookup(node.right);
int uid = trees.computeIfAbsent(serial, x -> t++);
if (count.get(uid) == 2) {
ans.add(nodes);
}
return uid;
}
}