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:
Input: root = [1,2,3,4], x = 4, y = 3
Output: false
Example 2:
Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true
Example 3:
Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false
Constraints:
The number of nodes in the tree will be between
2
and100
.Each node has a unique integer value from
1
to100
.
这道题定义了一种二叉树数的表兄弟结点，就是不属于同一个父结点，但是深度相同，现在给了两个结点值，问它们代表的结点是否是表兄弟结点。由于表兄弟结点一定是属于同一层的，所以可以使用二叉树的层序遍历，就像之前那道 Binary Tree Level Order Traversal 一样。这里额外需要两个布尔型变量 isX，isY 来记录x和y是否已经遍历到了。由于是层序遍历，所以 while 中需要有个 for 循环，在循环中，取出队首结点，然后看结点值是否等于x或y，是的话标记布尔变量。然后检查当前结点的左右子结点是否分别是x和y，是的话直接返回 false。否则将左右子结点排入队列之中，若存在的话。当前层遍历完了之后，检查 isX 和 isY 的值，若同时为 true，表示存在表兄弟结点，返回 true。若只有一个为 true 的话，说明二者不在同一层，直接返回 false，参见代码如下：
解法一：
class Solution {public:bool isCousins(TreeNode* root, int x, int y) {queue<TreeNode*> q{{root}};bool isX = false, isY = false;while (!q.empty()) {for (int i = q.size(); i > 0; --i) {TreeNode *cur = q.front(); q.pop();if (cur->val == x) isX = true;if (cur->val == y) isY = true;if (cur->left && cur->right) {int left = cur->left->val, right = cur->right->val;if ((left == x && right == y) || (left == y && right == x)) return false;}if (cur->left) q.push(cur->left);if (cur->right) q.push(cur->right);}if (isX && isY) return true;if (isX || isY) return false;}return false;}};
当然我们也可以用递归的方法来做，由于不能同时处理同一层的结点，就需要两个变量 depthX 和 depthY 来记录结点x和y的深度，同时再用个布尔变量 sameParent 来记录这两个结点是否有相同的父结点。在递归函数中，若当前结点 node 为空，直接返回。若当前结点值是x，则 depthX 赋值为当前深度 cur，同理，若当前结点值是y，则 depthY 赋值为当前深度 cur。然后检查当前结点的左右子结点是否分别是x和y，是的话 sameParent 标记为 true，最后分别对左右子结点调用递归即可，参见代码如下：
解法二：
class Solution {public:bool isCousins(TreeNode* root, int x, int y) {int depthX = 0, depthY = 0;bool sameParent = false;helper(root, x, y, 0, depthX, depthY, sameParent);return !sameParent && depthX == depthY;}void helper(TreeNode* node, int x, int y, int cur, int& depthX, int& depthY, bool& sameParent) {if (!node) return;if (node->val == x) depthX = cur;if (node->val == y) depthY = cur;if (node->left && node->right) {int left = node->left->val, right = node->right->val;if ((left == x && right == y) || (left == y && right == x)) sameParent = true;}helper(node->left, x, y, cur + 1, depthX, depthY, sameParent);helper(node->right, x, y, cur + 1, depthX, depthY, sameParent);}};
接下来这种方法是博主在 Contest 时想出来的，思路比较简单粗暴，就是建立每个结点和其深度跟父结点组成的 pair 对儿之间的映射，然后就可以直接用x和y去获得其深度和父结点进行比较即可，参见代码如下：
解法三：
class Solution {public:bool isCousins(TreeNode* root, int x, int y) {unordered_map<int, pair<int, int>> m;helper(root, 0, -1, m);return m[x].first == m[y].first && m[x].second != m[y].second;}void helper(TreeNode* node, int depth, int parent, unordered_map<int, pair<int, int>>& m) {if (!node) return;m[node->val] = {depth, parent};helper(node->left, depth + 1, node->val, m);helper(node->right, depth + 1, node->val, m);}};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/993
类似题目：
Binary Tree Level Order Traversal
参考资料：
https://leetcode.com/problems/cousins-in-binary-tree/
https://leetcode.com/problems/cousins-in-binary-tree/discuss/238536/My-Easy-Recursive-C%2B%2B-Solution
https://leetcode.com/problems/cousins-in-binary-tree/discuss/239376/Java-BFS-time-and-space-beat-100
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