Leetcode 993. Cousins in Binary Tree (二叉树遍历好题)

  1. Cousins in Binary Tree
    Easy
    3.9K
    193
    Companies
    Given the root of a binary tree with unique values and the values of two different nodes of the tree x and y, return true if the nodes corresponding to the values x and y in the tree are cousins, or false otherwise.

Two nodes of a binary tree are cousins if they have the same depth with different parents.

Note that in a binary tree, the root node is at the depth 0, and children of each depth k node are at the depth k + 1.

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 is in the range [2, 100].

1 <= Node.val <= 100

Each node has a unique value.

x != y

x and y are exist in the tree.

解法1:

cpp 复制代码
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isCousins(TreeNode* root, int x, int y) {
        if (root->val == x || root->val == y) return false;
        helper(root, x, y, 0);
          if (depthX == depthY && fatherX != fatherY) {
            return true;
        } else {
            return false;
        }
    }
private:
    TreeNode *fatherX = NULL, *fatherY = NULL;
    int depthX = 0, depthY = 0;
    void helper(TreeNode *root, int x, int y, int depth) {
        if (!root || (fatherX && fatherY)) return;
        if (!fatherX) {
            if ((root->left && root->left->val == x) || 
                (root->right && root->right->val == x)) {
                depthX = depth + 1;
                fatherX = root;
            }
        }  
        if (!fatherY) {
            if ((root->left && root->left->val == y) || 
                (root->right && root->right->val == y)) {
                depthY = depth + 1;
                fatherY = root;
            }
        }
        helper(root->left, x, y, depth + 1);
        helper(root->right, x, y, depth + 1);
    }
};

二刷:

cpp 复制代码
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isCousins(TreeNode* root, int x, int y) {
        helper(root, x, y, 0, NULL);
        if (depthX == depthY && fatherX != fatherY) return true;
        return false;
    }
private:
    int depthX = 0, depthY = 0;
    TreeNode *fatherX = NULL, *fatherY = NULL;
    void helper(TreeNode *root, int x, int y, int depth, TreeNode *father) {
        if (!root) return;
        if (root->val == x) {
            fatherX = father;
            depthX = depth;
        }
        if (root->val == y) {
            fatherY = father;
            depthY = depth;
        }
        helper(root->left, x, y, depth + 1, root);
        helper(root->right, x, y, depth + 1, root);
    }
};
相关推荐
范纹杉想快点毕业11 分钟前
欧几里得算法与扩展欧几里得算法,C语言编程实现(零基础全解析)
运维·c语言·单片机·嵌入式硬件·算法
f***241112 分钟前
Bug悬案:技术侦探的破案指南
算法·bug
Swift社区14 分钟前
LeetCode 472 连接词
算法·leetcode·职场和发展
Dream it possible!15 分钟前
LeetCode 面试经典 150_二分查找_搜索旋转排序数组(114_33_C++_中等)
c++·leetcode·面试
CoovallyAIHub21 分钟前
YOLO-Maste开源:首个MoE加速加速实时检测,推理提速17.8%!
深度学习·算法·计算机视觉
清铎25 分钟前
leetcode_day13_普通数组_《绝境求生》
数据结构·算法
2301_8008951032 分钟前
hh的蓝桥杯每日一题(二分)--立定跳远
职场和发展·蓝桥杯
hetao173383735 分钟前
2026-01-09~12 hetao1733837 的刷题笔记
c++·笔记·算法
过河卒_zh15667661 小时前
情感型AI被“立规矩”,AI陪伴时代进入下半场
人工智能·算法·aigc·生成式人工智能·算法备案
wefg11 小时前
【算法】动态规划
算法·动态规划