Problem solution 993 “Cousins in Binary Tree” from LeetCode.
Problem
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
notes:
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.
Solution
Breadth First Search
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/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isCousins(root *TreeNode, x int, y int) bool {
q := [][]*TreeNode{}
if root == nil {
return true
}
if root.Left!=nil && root.Right!=nil {
if (root.Left.Val == x || root.Left.Val == y) &&
(root.Right.Val == x || root.Right.Val == y) {
return false
}
}
q = append(q, []*TreeNode{root.Left, root.Right})
for k:=0;k<len(q); {
d := []*TreeNode{}
c:=0
for idx:=range q[len(q)-1] {
if q[len(q)-1][idx] == nil {
continue
}
if (q[len(q)-1][idx].Val == x) || (q[len(q)-1][idx].Val == y){
c++
}
if c == 2 {
return true
}
r := q[len(q)-1][idx]
if (r.Left != nil && r.Right!= nil) {
if (r.Left.Val == x || r.Left.Val == y) &&
(r.Right.Val == x || r.Right.Val == y) {
return false
}
}
if q[len(q)-1][idx].Left != nil {
d = append(d,q[len(q)-1][idx].Left)
}
if q[len(q)-1][idx].Right != nil {
d = append(d,q[len(q)-1][idx].Right)
}
}
if len(d)>0 {
q = append(q, d)
}
k++
}
return false
}
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