DFS
class Solution {
int x, y;
int depthX, depthY;
TreeNode fatherX, fatherY;
public boolean isCousins(TreeNode root, int x, int y) {
this.x = x;
this.y = y;
getNodeDepthAndHisFather(root, null, 0);
return (depthX == depthY) && (fatherX != fatherY);
}
void getNodeDepthAndHisFather(TreeNode root, TreeNode father, int depth){
if(root == null) {return;}
if(root.val == x) {
this.depthX = depth;
this.fatherX = father;
}
if(root.val == y) {
this.depthY = depth;
this.fatherY = father;
}
getNodeDepthAndHisFather(root.left, root, depth+1);
getNodeDepthAndHisFather(root.right, root, depth+1);
}
}
BFS
public boolean isCousins(TreeNode root, int x, int y) {
//兩個隊列一個存放樹的節點,一個存放節點對應的值
Queue<TreeNode> queue = new LinkedList<>();
Queue<Integer> value = new LinkedList<>();
queue.add(root);
value.add(root.val);
//如果隊列不為空,說明樹的節點沒有周遊完,就繼續周遊
while (!queue.isEmpty()) {
//BFS是從上到下一層一層的列印,levelSize表示
//目前層的節點個數
int levelSize = queue.size();
for (int i = 0; i < levelSize; i++) {
//節點和節點值同時出隊
TreeNode poll = queue.poll();
value.poll();
//首先判斷x和y是否是兄弟節點的值,也就是判斷他們的父節點
//是否是同一個
if (poll.left != null && poll.right != null) {
if ((poll.left.val == x && poll.right.val == y) ||
(poll.left.val == y && poll.right.val == x)) {
return false;
}
}
//左子節點不為空加入到隊列中
if (poll.left != null) {
queue.offer(poll.left);
value.offer(poll.left.val);
}
//右子節點不為空加入到隊列中
if (poll.right != null) {
queue.offer(poll.right);
value.offer(poll.right.val);
}
}
//判斷目前層是否包含這兩個節點的值,如果包含就是堂兄弟節點
if (value.contains(x) && value.contains(y))
return true;
}
return false;
}