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Copy pathDay7. CousinsInBinaryTree.cpp
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Day7. CousinsInBinaryTree.cpp
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/*
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:
https://assets.leetcode.com/uploads/2019/02/12/q1248-01.png
Input: root = [1,2,3,4], x = 4, y = 3
Output: false
Example 2:
https://assets.leetcode.com/uploads/2019/02/12/q1248-02.png
Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true
Example 3:
https://assets.leetcode.com/uploads/2019/02/13/q1248-03.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.
*/
//Solution approach: use Breadth First search checking if x and y have same parent or not if yes return false.
// otherwise if x and y are in same level return true.
class Solution {
public:
bool isCousins(TreeNode* root, int x, int y) {
if(root==NULL) return 0;
queue<TreeNode*> q;
TreeNode* i;
q.push(root);
int count=0;
while(!q.empty()){
int n=q.size();
count=0;
while(n){
i=q.front();
if(i->val==x || i->val==y){
count++;
}
q.pop();
if(i->left!=NULL && i->right!=NULL){
if(i->left->val==x && i->right->val==y) return false;
if(i->left->val==y && i->right->val==x) return false;
}
if(i->left) q.push(i->left);
if(i->right) q.push(i->right);
n--;
}
if(count==2){
return true;
}
}
return false;
}
};