LeetCode 993. Cousins in Binary Tree(判断结点是否为Cousin)
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:
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
思路:
求解x,y的深度和父亲结点,如果深度一样,父亲结点不同,就是true;否则,就是false。
/**
* 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:
TreeNode *getDepth(TreeNode* root, int val, int depth, int &level)
{
if (!root) return nullptr; //找到val的父亲,并设置val的depth
if ((root->left && root->left->val == val) || (root->right && root->right->val == val))
{
level = depth;
return root;
} TreeNode *left = getDepth(root->left, val, depth + , level);
if (left) return left; TreeNode *right = getDepth(root->right, val, depth + , level);
if (right) return right; return nullptr;
} bool isCousins(TreeNode *root, int x, int y)
{
int x_depth = -, y_depth = -;
TreeNode *x_parent = getDepth(root, x, , x_depth);
TreeNode *y_parent = getDepth(root, y, , y_depth);
if (x_depth == y_depth && x_parent != y_parent)
{
return true;
}
return false;
}
};
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