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
Note:
- The number of nodes in the tree will be between
2
and100
. - Each node has a unique integer value from
1
to100
.
-----------------------------------------------------------------------------------------
额,这个题,怎么说的,弄懂了方法,就会发现这个题是比较简单的,是easy题是有道理的。要好好扎实BFS的基础。
这个题就是先把每个结点的父节点和它的位置求出,然后再比较,可以用hash。
C++代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isCousins(TreeNode* root, int x, int y) {
queue<pair<TreeNode*,TreeNode*> > q;
q.push(make_pair(root,nullptr));
int depth = ;
unordered_map<int,pair<TreeNode*,int> > mp; //TreeNode*指的是这个结点的父结点。第二个int指的是深度。
while(!q.empty()){
for(int i = q.size(); i > ; i--){
auto t = q.front();
q.pop();
mp[t.first->val] = make_pair(t.second,depth);
if(t.first->left) q.push(make_pair(t.first->left,t.first));
if(t.first->right) q.push(make_pair(t.first->right,t.first));
}
depth++;
}
auto tx = mp[x],ty = mp[y];
return tx.first != ty.first && tx.second == ty.second;
}
};