993. Cousins in Binary Tree*

发布于:2021-10-14 10:48:52

993. Cousins in Binary Tree*

https://leetcode.com/problems/cousins-in-binary-tree/


题目描述

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 and 100.Each node has a unique integer value from 1 to 100.
C++ 实现 1

这道题感觉用层序遍历更为方便. 访问每一层的时候, 需要做如下判断:


如果该层只有一个节点, 而且还满足条件, 那么返回 false. 我这里的 “满足条件” 指的是该节点的值刚好等于 x 或者 y. 以下描述中的满足条件都是这个意思;如果当前层有多个节点, 对于某个具体的节点 r 如果它的左右孩子满足条件, 当然不是 cousins, 需要返回 false; 那么如何表示某个节点左右孩子满足条件呢? 下面代码中我用 num 来判断某个节点 r 的两个孩子是否同时满足条件, 如果同时满足, 那么返回 false. 另外使用 count 来判断这一层的所有节点是否满足条件, 如果满足, 则返回 true (if (count == 2) return true;).如果当前层的所有节点遍历完, 发现 count == 1, 此时说明, 就算下一层存在节点满足条件, 也不可能是 cousins 了, 因为两个节点不在同一层, 所以要返回 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:
bool isCousins(TreeNode* root, int x, int y) {
unordered_set record{x, y};
queue q;
q.push(root);
while (!q.empty()) {
auto size = q.size();
if (size == 1 && record.count(q.front()->val)) return false;
int count = 0;
while (size --) {
int num = 0; // num 用来记录当前节点的左右孩子是否满足条件
auto r = q.front();
q.pop();
if (r->left) {
q.push(r->left);
if (record.count(r->left->val)) {
count ++;
num ++;
}
}
if (r->right) {
q.push(r->right);
// 同一个节点的左右孩子如果满足条件, 那么就不是 cousins
if (record.count(r->right->val)) {
if (num == 1) return false;
count ++;
}
}
}
// 当前层如果只有一个节点满足条件, 那么当然不是 cousins
if (count == 1) return false;
if (count == 2) return true;
}
return false;
}
};

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