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

Cousins in Binary Tree
Easy
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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);
    }
};