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[leetcode] problem 993 – cousins in binary tree

admin 11月 13, 2020 0

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Example

No.1

EbRMX4.png

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

Output: false

No.2

EbR37R.png

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

Output: true

No.3

EbRhuQ.png

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

Output: false

Note

  1. The number of nodes in the tree will be between 2 and 100.
  2. Each node has a unique integer value from 1 to 100.

Code

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public class  {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}
 
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public boolean isCousins(TreeNode root, int x, int y) {
    Queue<TreeNode> queue = new LinkedList<>();
    queue.offer(root);

    while (!queue.isEmpty()) {
        int size = queue.size();
        int xParent = 0;
        int yParent = 0;

        for (int i = 0; i < size; i++) {
            TreeNode node = queue.poll();

            if (node.left != null) {
                queue.offer(node.left);

                if (node.left.val == x)
                    xParent = node.val;
                if (node.left.val == y)
                    yParent = node.val;
            }

            if (node.right != null) {
                queue.offer(node.right);

                if (node.right.val == x)
                    xParent = node.val;
                if (node.right.val == y)
                    yParent = node.val;
            }
        }

        if (xParent != 0 && yParent != 0)
            return xParent != yParent;
        else if (xParent != 0 || yParent != 0)
            return false;
    }

    return false;
}
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