993. Cousins in Binary Tree

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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:

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Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

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Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

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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.

Hints/Notes

  • N/A

Solution

Language: C++

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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:
    int x_;
    int y_;
    int depthX = -1;
    int depthY = -1;
    TreeNode* parentX;
    TreeNode* parentY;

    bool isCousins(TreeNode* root, int x, int y) {
        x_ = x;
        y_ = y;
        dfs(root, 0, nullptr);
        return depthX == depthY && parentX != parentY;
    }

    void dfs(TreeNode* root, int depth, TreeNode* parent) {
        if (!root) {
            return;
        }

        if (depthX != -1 && depthY != -1) {
            return;
        }

        if (root->val == x_) {
            depthX = depth;
            parentX = parent;
        }

        if (root->val == y_) {
            depthY = depth;
            parentY = parent;
        }

        dfs(root->left, depth + 1, root);
        dfs(root->right, depth + 1, root);
    }
};