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 != yx and y are exist in the tree.
Solutions#
Solution 1#
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| # 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:
q = deque([(root, None)])
d = 0
p1 = p2 = None
d1 = d2 = 0
while q:
for _ in range(len(q)):
node, fa = q.popleft()
if node.val == x:
p1, d1 = fa, d
if node.val == y:
p2, d2 = fa, d
if node.left:
q.append((node.left, node))
if node.right:
q.append((node.right, node))
d += 1
return p1 != p2 and d1 == d2
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| /**
* 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 {
public boolean isCousins(TreeNode root, int x, int y) {
TreeNode p1 = null, p2 = null;
int d1 = 0, d2 = 0;
Deque<TreeNode[]> q = new ArrayDeque<>();
q.offer(new TreeNode[] {root, null});
int d = 0;
while (!q.isEmpty()) {
for (int n = q.size(); n > 0; --n) {
var p = q.poll();
TreeNode node = p[0], fa = p[1];
if (node.val == x) {
p1 = fa;
d1 = d;
}
if (node.val == y) {
p2 = fa;
d2 = d;
}
if (node.left != null) {
q.offer(new TreeNode[] {node.left, node});
}
if (node.right != null) {
q.offer(new TreeNode[] {node.right, node});
}
}
++d;
}
return p1 != p2 && d1 == d2;
}
}
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| /**
* 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 = nullptr;
TreeNode* p2 = nullptr;
int d1 = 0, d2 = 0;
queue<pair<TreeNode*, TreeNode*>> q;
q.emplace(root, nullptr);
int d = 0;
while (!q.empty()) {
for (int n = q.size(); n; --n) {
auto [node, fa] = q.front();
q.pop();
if (node->val == x) {
p1 = fa;
d1 = d;
}
if (node->val == y) {
p2 = fa;
d2 = d;
}
if (node->left) {
q.emplace(node->left, node);
}
if (node->right) {
q.emplace(node->right, node);
}
}
++d;
}
return p1 != p2 && d1 == d2;
}
};
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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 {
type pair struct{ node, fa *TreeNode }
q := []pair{pair{root, nil}}
var p1, p2 *TreeNode
var d, d1, d2 int
for len(q) > 0 {
for n := len(q); n > 0; n-- {
p := q[0]
q = q[1:]
node, fa := p.node, p.fa
if node.Val == x {
p1, d1 = fa, d
}
if node.Val == y {
p2, d2 = fa, d
}
if node.Left != nil {
q = append(q, pair{node.Left, node})
}
if node.Right != nil {
q = append(q, pair{node.Right, node})
}
}
d++
}
return p1 != p2 && d1 == d2
}
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Solution 2#
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| # 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]
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| /**
* 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);
}
}
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| /**
* 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;
}
};
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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 {
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
}
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