652. 寻找重复的子树

English Version

题目描述

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。

解法

后序遍历，序列化每个子树，用哈希表判断序列化的字符串出现次数是否等于 2，若是，说明这棵子树重复。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findDuplicateSubtrees(self, root: Optional[TreeNode]) -> List[Optional[TreeNode]]:
        def dfs(root):
            if root is None:
                return '#'
            v = f'{root.val},{dfs(root.left)},{dfs(root.right)}'
            counter[v] += 1
            if counter[v] == 2:
                ans.append(root)
            return v

        ans = []
        counter = Counter()
        dfs(root)
        return ans

Java

/**
 * 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 {
    private Map<String, Integer> counter;
    private List<TreeNode> ans;

    public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
        counter = new HashMap<>();
        ans = new ArrayList<>();
        dfs(root);
        return ans;
    }

    private String dfs(TreeNode root) {
        if (root == null) {
            return "#";
        }
        String v = root.val + "," + dfs(root.left) + "," + dfs(root.right);
        counter.put(v, counter.getOrDefault(v, 0) + 1);
        if (counter.get(v) == 2) {
            ans.add(root);
        }
        return v;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    unordered_map<string, int> counter;
    vector<TreeNode*> ans;

    vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
        dfs(root);
        return ans;
    }

    string dfs(TreeNode* root) {
        if (!root) return "#";
        string v = to_string(root->val) + "," + dfs(root->left) + "," + dfs(root->right);
        ++counter[v];
        if (counter[v] == 2) ans.push_back(root);
        return v;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findDuplicateSubtrees(root *TreeNode) []*TreeNode {
	var ans []*TreeNode
	counter := make(map[string]int)
	var dfs func(root *TreeNode) string
	dfs = func(root *TreeNode) string {
		if root == nil {
			return "#"
		}
		v := strconv.Itoa(root.Val) + "," + dfs(root.Left) + "," + dfs(root.Right)
		counter[v]++
		if counter[v] == 2 {
			ans = append(ans, root)
		}
		return v
	}
	dfs(root)
	return ans
}

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