Given two binary trees original and cloned and given a reference to a node target in the original tree.

The cloned tree is a copy of the original tree.

Return a reference to the same node in the cloned tree.

Note that you are not allowed to change any of the two trees or the target node and the answer must be a reference to a node in the cloned tree.

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Example 1:

Input: tree = [7,4,3,null,null,6,19], target = 3
Output: 3
Explanation: In all examples the original and cloned trees are shown. The target node is a green node from the original tree. The answer is the yellow node from the cloned tree.


Example 2:

Input: tree = [7], target =  7
Output: 7


Example 3:

Input: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
Output: 4


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Constraints:


	The number of nodes in the tree is in the range [1, 104].
	The values of the nodes of the tree are unique.
	target node is a node from the original tree and is not null.


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Follow up: Could you solve the problem if repeated values on the tree are allowed?

class Solution:
    def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode:
        def DFS(node1,node2):
            if node1==target:
                return node2
            if node1 and node1.left is None and node1.right is None:
                return

            res1 = DFS(node1.left,node2.left) if node1 else None
            if res1 is not None:
                return res1
            res2 = DFS(node1.right,node2.right) if node1 else None
            if res2 is not None:
                return res2
        res=DFS(original,cloned)
        return res
class Solution {
    public final TreeNode getTargetCopy(final TreeNode original, final TreeNode cloned, final TreeNode target) {
        TreeNode[] ref = new TreeNode[]{null};
        dfs(cloned, target, ref);
        return ref[0];
    }
    public static void dfs (TreeNode root, TreeNode target, TreeNode[] ref) {
        if (root == null) return;
        if (root.val == target.val) {
            ref[0] = root;
            return;
        } else {
            dfs(root.left, target, ref);
            dfs(root.right, target, ref);
        }
    }
}
class Solution {
public:
    TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
        if(original == target || original == NULL)
            return cloned;
        TreeNode* found_left = getTargetCopy(original->left, cloned->left, target);
        TreeNode* found_right = getTargetCopy(original->right, cloned->right, target);
        if(found_left)
            return found_left;
        else
            return found_right;
    }
};
var getTargetCopy = function(original, cloned, target) {
    
    if( original == null ){
        
        // Base case aka stop condition
        // empty tree or empty node
        return null;
    }
    
    // General cases
    if( original == target ){
       
        // current original node is target, so is cloned
        return cloned;
    }
    
    // Either left subtree has target, or right subtree has target
    return getTargetCopy(original.left, cloned.left, target) || 
           getTargetCopy(original.right, cloned.right, target);
    
};
Find a Corresponding Node of a Binary Tree in a Clone of That Tree