Given the root of 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 the same node values.
Example 1:
Input: root = [1,2,3,4,null,2,4,null,null,4]
Output: [[2,4],[4]]
Example 2:
Input: root = [2,1,1]
Output: [[1]]
Example 3:
Input: root = [2,2,2,3,null,3,null]
Output: [[2,3],[3]]
Constraints:
- The number of the nodes in the tree will be in the range
[1, 5000] -200 <= Node.val <= 200
package main
import "strconv"
// Definition for a binary tree node.
type TreeNode struct {
Val int
Left *TreeNode
Right *TreeNode
}
func findDuplicateSubtrees(root *TreeNode) []*TreeNode {
// to store the hash value of subtrees
hashMap := make(map[string]int)
// to store the result
res := make([]*TreeNode, 0)
// traverse the tree
traverse(root, hashMap, &res)
return res
}
// traverse the tree
func traverse(root *TreeNode, hashMap map[string]int, res *[]*TreeNode) string {
// if the root is nil, return "#"
if root == nil {
return "#"
}
// traverse the left and right subtree
left := traverse(root.Left, hashMap, res)
right := traverse(root.Right, hashMap, res)
// get the hash value of the subtree
subTree := left + "," + right + "," + strconv.Itoa(root.Val)
// if the hash value of the subtree is 1, it means that the subtree is duplicated
if hashMap[subTree] == 1 {
*res = append(*res, root)
}
// add the hash value of the subtree to the hashMap
hashMap[subTree]++
return subTree
}