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:
\
2 3
/ / \
The following are two duplicate subtrees:
2
/
4
and
4
Therefore, you need to return above trees’ root in the form of a list.
思路：
Intuition
We can serialize each subtree. For example, the tree
\
2 3
/ \
can be represented as the serialization 1,2,#,#,3,4,#,#,5,#,#, which is a unique representation of the tree.
Algorithm
Perform a depth-first search, where the recursive function returns the serialization of the tree. At each node, record the result in a map, and analyze the map after to determine duplicate subtrees.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
Map<String, Integer> count;
List<TreeNode> ans;
public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
count = new HashMap();
ans = new ArrayList();
collect(root);
return ans;
}
public String collect(TreeNode node) {
if (node == null)
return "#";
String serial = node.val + "," + collect(node.left) + "," + collect(node.right);
count.put(serial, count.getOrDefault(serial, 0) + 1);
if (count.get(serial) == 2)
ans.add(node);
return
