给定一棵二叉树，返回所有重复的子树。对于同一类的重复子树，你只需要返回其中任意一棵的根结点即可。

两棵树重复是指它们具有相同的结构以及相同的结点值。

示例 1：

    1
   / \
  2   3
 /   / \
4   2   4
   /
  4

下面是两个重复的子树：

  2
 /
4

和

4

因此，你需要以列表的形式返回上述重复子树的根结点。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-duplicate-subtrees
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

深度优先搜索

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {

    private List<TreeNode> ans = new ArrayList<>();

    private Map<String, Integer> cntMap = new HashMap<>();

    private String solve(TreeNode root) {
        if (root == null) {
            return "#";
        }
        String series = root.val + "+" + solve(root.left) + "+" + solve(root.right);
        Integer cnt = cntMap.getOrDefault(series, 0);
        if (cnt == 1) {
            ans.add(root);
        }
        cntMap.put(series, cnt + 1);
        return series;
    }

    public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
        solve(root);
        return ans;
    }

}

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;
    }
}

唯一标识符号

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

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;
    }

    public int lookup(TreeNode node) {
        if (node == null) {
            return 0;
        }
        String serial = node.val + "," + lookup(node.left) + "," + lookup(node.right);
        int uid = trees.computeIfAbsent(serial, x -> t++);
        count.put(uid, count.getOrDefault(uid, 0) + 1);
        if (count.get(uid) == 2)
            ans.add(node);
        return uid;
    }
}


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;
    }
}