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993. Cousins in Binary Tree

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 != y
  • x 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)])
        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

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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) {
        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;
    }
}

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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) {
        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