# Leetcode: Cousins in Binary Tree

Cousins in Binary Tree

Similar Problems:

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

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

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

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
```

Note:

1. The number of nodes in the tree will be between 2 and 100.
2. Each node has a unique integer value from 1 to 100.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```// Blog link: https://code.dennyzhang.com/cousins-in-binary-tree
// Basic Ideas: dfs, bfs
// track level + parent node
// Complexity: Time O(n), Space O(1)
/**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
type node struct {
parent *TreeNode
level int
}

func dfs(root *TreeNode, level int, parent *TreeNode, x int) *node {
if root == nil {
return &node{nil, -1}
}
if root.Val == x {
return &node{parent, level}
}
node := dfs(root.Left, level+1, root, x)
if node.parent != nil {
return node
} else {
return dfs(root.Right, level+1, root, x)
}
}
func isCousins(root *TreeNode, x int, y int) bool {
node1, node2 := dfs(root, 0, nil, x), dfs(root, 0, nil, y)
return node1.parent != node2.parent && node1.level == node2.level
}
```

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