# 993. Cousins in Binary Tree

## Description

Given the root of a binary tree with unique values and the values of two different nodes of the tree x and y, return true if the nodes corresponding to the values x and y in the tree are cousins, or false otherwise.

Two nodes of a binary tree are cousins if they have the same depth with different parents.

Note that in a binary tree, the root node is at the depth 0, and children of each depth k node are at the depth k + 1.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false


Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true


Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false


Constraints:

• The number of nodes in the tree is in the range [2, 100].
• 1 <= Node.val <= 100
• Each node has a unique value.
• x != y
• x and y are exist in the tree.

## Solutions

BFS or DFS.

• /**
* 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 int x, y;
private TreeNode p1, p2;
private int d1, d2;

public boolean isCousins(TreeNode root, int x, int y) {
this.x = x;
this.y = y;
dfs(root, null, 0);
return p1 != p2 && d1 == d2;
}

private void dfs(TreeNode root, TreeNode p, int d) {
if (root == null) {
return;
}
if (root.val == x) {
p1 = p;
d1 = d;
}
if (root.val == y) {
p2 = p;
d2 = d;
}
dfs(root.left, root, d + 1);
dfs(root.right, root, d + 1);
}
}

• /**
* 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:
bool isCousins(TreeNode* root, int x, int y) {
TreeNode *p1, *p2;
int d1, d2;
function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* fa, int d) {
if (!root) {
return;
}
if (root->val == x) {
p1 = fa;
d1 = d;
}
if (root->val == y) {
p2 = fa;
d2 = d;
}
dfs(root->left, root, d + 1);
dfs(root->right, root, d + 1);
};
dfs(root, nullptr, 0);
return p1 != p2 && d1 == d2;
}
};

• # 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 isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
def dfs(root, fa, d):
if root is None:
return
if root.val == x:
t[0] = (fa, d)
if root.val == y:
t[1] = (fa, d)
dfs(root.left, root, d + 1)
dfs(root.right, root, d + 1)

t = [None, None]
dfs(root, None, 0)
return t[0][0] != t[1][0] and t[0][1] == t[1][1]


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func isCousins(root *TreeNode, x int, y int) bool {
var p1, p2 *TreeNode
var d1, d2 int
var dfs func(*TreeNode, *TreeNode, int)
dfs = func(root *TreeNode, fa *TreeNode, d int) {
if root == nil {
return
}
if root.Val == x {
p1, d1 = fa, d
}
if root.Val == y {
p2, d2 = fa, d
}
dfs(root.Left, root, d+1)
dfs(root.Right, root, d+1)
}
dfs(root, nil, 0)
return p1 != p2 && d1 == d2
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function isCousins(root: TreeNode | null, x: number, y: number): boolean {
let [d1, d2] = [0, 0];
let [p1, p2] = [null, null];
const q: [TreeNode, TreeNode][] = [[root, null]];
for (let depth = 0; q.length > 0; ++depth) {
const t: [TreeNode, TreeNode][] = [];
for (const [node, parent] of q) {
if (node.val === x) {
[d1, p1] = [depth, parent];
} else if (node.val === y) {
[d2, p2] = [depth, parent];
}
if (node.left) {
t.push([node.left, node]);
}
if (node.right) {
t.push([node.right, node]);
}
}
q.splice(0, q.length, ...t);
}
return d1 === d2 && p1 !== p2;
}