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_993.java
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_993.java
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package com.fishercoder.solutions;
import com.fishercoder.common.classes.TreeNode;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;
/**
* 993. Cousins in Binary Tree
*
* 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 1:
* 1
* / \
* 2 3
* /
* 4
*
* Input: root = [1,2,3,4], x = 4, y = 3
* Output: false
*
* Example 2:
* 1
* / \
* 2 3
* \ \
* 4 5
*
* Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
* Output: true
*
* Example 3:
* 1
* / \
* 2 3
* \
* 4
*
* Input: root = [1,2,3,null,4], x = 2, y = 3
* Output: false
*
*
* Note:
*
* The number of nodes in the tree will be between 2 and 100.
* Each node has a unique integer value from 1 to 100.
*/
public class _993 {
public static class Solution1 {
public boolean isCousins(TreeNode root, int x, int y) {
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode current = queue.poll();
if (current.left != null) {
queue.offer(current.left);
}
if (current.right != null) {
queue.offer(current.right);
}
if (current.left != null && current.right != null) {
if (current.left.val == x && current.right.val == y
|| current.left.val == y && current.right.val == x) {
return false;
}
}
}
if (checkQueue(queue, x, y)) {
return true;
}
}
return false;
}
private boolean checkQueue(Queue<TreeNode> queue, int x, int y) {
Set<Integer> set = new HashSet<>();
Queue<TreeNode> tmp = new LinkedList<>(queue);
while (!tmp.isEmpty()) {
set.add(tmp.poll().val);
}
return set.contains(x) && set.contains(y);
}
}
}