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
PythonJavaC++GoTypeScript
# 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)])
depth = 0
p1 = p2 = None
d1 = d2 = None
while q:
for _ in range(len(q)):
node, parent = q.popleft()
if node.val == x:
p1, d1 = parent, depth
elif node.val == y:
p2, d2 = parent, depth
if node.left:
q.append((node.left, node))
if node.right:
q.append((node.right, node))
depth += 1
return p1 != p2 and d1 == d2(code-box)
/**
* 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) {
Deque<TreeNode[]> q = new ArrayDeque<>();
q.offer(new TreeNode[] {root, null});
int d1 = 0, d2 = 0;
TreeNode p1 = null, p2 = null;
for (int depth = 0; !q.isEmpty(); ++depth) {
for (int n = q.size(); n > 0; --n) {
TreeNode[] t = q.poll();
TreeNode node = t[0], parent = t[1];
if (node.val == x) {
d1 = depth;
p1 = parent;
} else if (node.val == y) {
d2 = depth;
p2 = parent;
}
if (node.left != null) {
q.offer(new TreeNode[] {node.left, node});
}
if (node.right != null) {
q.offer(new TreeNode[] {node.right, node});
}
}
}
return p1 != p2 && d1 == d2;
}
}(code-box)
/**
* 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) {
queue<pair<TreeNode*, TreeNode*>> q;
q.push({root, nullptr});
int d1 = 0, d2 = 0;
TreeNode *p1 = nullptr, *p2 = nullptr;
for (int depth = 0; q.size(); ++depth) {
for (int n = q.size(); n; --n) {
auto [node, parent] = q.front();
q.pop();
if (node->val == x) {
d1 = depth;
p1 = parent;
} else if (node->val == y) {
d2 = depth;
p2 = parent;
}
if (node->left) {
q.push({node->left, node});
}
if (node->right) {
q.push({node->right, node});
}
}
}
return d1 == d2 && p1 != p2;
}
};(code-box)
/**
* 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, parent *TreeNode }
var d1, d2 int
var p1, p2 *TreeNode
q := []pair{{root, nil}}
for depth := 0; len(q) > 0; depth++ {
for n := len(q); n > 0; n-- {
node, parent := q[0].node, q[0].parent
q = q[1:]
if node.Val == x {
d1, p1 = depth, parent
} else if node.Val == y {
d2, p2 = depth, parent
}
if node.Left != nil {
q = append(q, pair{node.Left, node})
}
if node.Right != nil {
q = append(q, pair{node.Right, node})
}
}
}
return d1 == d2 && p1 != p2
}(code-box)
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function isCousins(root: TreeNode | null, x: number, y: number): boolean {
let [d1, d2] = [0, 0];
let [p1, p2] = [null, null];
const q: [TreeNode, TreeNode][] = [[root, null]];
for (let depth = 0; q.length > 0; ++depth) {
const t: [TreeNode, TreeNode][] = [];
for (const [node, parent] of q) {
if (node.val === x) {
[d1, p1] = [depth, parent];
} else if (node.val === y) {
[d2, p2] = [depth, parent];
}
if (node.left) {
t.push([node.left, node]);
}
if (node.right) {
t.push([node.right, node]);
}
}
q.splice(0, q.length, ...t);
}
return d1 === d2 && p1 !== p2;
}(code-box)
Solution 2
PythonJavaC++GoTypeScript
# 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, parent, depth):
if root is None:
return
if root.val == x:
st[0] = (parent, depth)
elif root.val == y:
st[1] = (parent, depth)
dfs(root.left, root, depth + 1)
dfs(root.right, root, depth + 1)
st = [None, None]
dfs(root, None, 0)
return st[0][0] != st[1][0] and st[0][1] == st[1][1](code-box)
/**
* 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 int d1, d2;
private TreeNode p1, p2;
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 parent, int depth) {
if (root == null) {
return;
}
if (root.val == x) {
d1 = depth;
p1 = parent;
} else if (root.val == y) {
d2 = depth;
p2 = parent;
}
dfs(root.left, root, depth + 1);
dfs(root.right, root, depth + 1);
}
}(code-box)
/**
* 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) {
int d1, d2;
TreeNode* p1;
TreeNode* p2;
function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* parent, int depth) {
if (!root) {
return;
}
if (root->val == x) {
d1 = depth;
p1 = parent;
} else if (root->val == y) {
d2 = depth;
p2 = parent;
}
dfs(root->left, root, depth + 1);
dfs(root->right, root, depth + 1);
};
dfs(root, nullptr, 0);
return p1 != p2 && d1 == d2;
}
};(code-box)
/**
* 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 d1, d2 int
var p1, p2 *TreeNode
var dfs func(root, parent *TreeNode, depth int)
dfs = func(root, parent *TreeNode, depth int) {
if root == nil {
return
}
if root.Val == x {
d1, p1 = depth, parent
} else if root.Val == y {
d2, p2 = depth, parent
}
dfs(root.Left, root, depth+1)
dfs(root.Right, root, depth+1)
}
dfs(root, nil, 0)
return d1 == d2 && p1 != p2
}(code-box)
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function isCousins(root: TreeNode | null, x: number, y: number): boolean {
let [d1, d2, p1, p2] = [0, 0, null, null];
const dfs = (root: TreeNode | null, parent: TreeNode | null, depth: number) => {
if (!root) {
return;
}
if (root.val === x) {
[d1, p1] = [depth, parent];
} else if (root.val === y) {
[d2, p2] = [depth, parent];
}
dfs(root.left, root, depth + 1);
dfs(root.right, root, depth + 1);
};
dfs(root, null, 0);
return d1 === d2 && p1 !== p2;
}(code-box)
