Cousins in Binary Tree Problem & Solution
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.
See the cousins in binary tree problem on LeetCode.
C++ Solution
#pragma GCC optimize("Ofast")
#pragma GCC optimization("max-inline-insns-recursive-auto")
static const int _=[](){std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();
/**
* 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) {
metadata left = getMetadata(root, x, 0);
metadata right = getMetadata(root, y, 0);
return left.depth == right.depth &&
left.parent != right.parent;
}
private:
struct metadata {
TreeNode* parent;
int depth;
};
metadata getMetadata(TreeNode* root, int val, int depth) {
if (root == nullptr) {
return {nullptr, 0};
}
if (root->left && root->left->val == val) {
return {root, depth + 1};
}
if (root->right && root->right->val == val) {
return {root, depth + 1};
}
metadata left = getMetadata(root->left, val, depth + 1);
if (left.parent) {
return left;
} else {
return getMetadata(root->right, val, depth + 1);
}
}
};