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:

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.

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
}