993. Cousins in Binary Tree

Easy
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:
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​
Input: root = [1,2,3,4], x = 4, y = 3
Output:
false
Example 2:
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​
Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output:
true
Example 3:
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​
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.

解題

Runtime: 0 ms, faster than 100%
Memory Usage: 2.5 MB, less than 82.86%
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isCousins(root *TreeNode, x int, y int) bool {
if root == nil {
return false
}
​
xLength := -1
xPre := root
yLength := -1
yPre := root
​
var findDepth func(*TreeNode, *TreeNode, int)
findDepth = func(root *TreeNode, pre *TreeNode, depth int) {
if root == nil {
return
}
​
if root.Val == x {
xLength = depth
xPre = pre
}
​
if root.Val == y {
yLength = depth
yPre = pre
}
​
findDepth(root.Left,root ,depth + 1)
findDepth(root.Right,root ,depth + 1)
}
​
findDepth(root, nil, 0)
​
return xLength == yLength && xPre != yPre
}