LeetCode 每日一题 872.叶子相似的树

---

问题描述

简要思路

昨天的每日一题，因种种原因忘了记录上传。
简单题，经典树的递归，两份代码是两种写法，应该算不上两种解法。写法2的三个if条件在树的题目还是蛮常用的。

代码

写法1：

/**
 * 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:
    void findLeaf(TreeNode* root, vector<int>& leaf){
    
    
        if(root == nullptr) return;
        if(root->left == nullptr && root->right == nullptr)
         leaf.push_back(root->val);
        findLeaf(root->left,leaf);
        findLeaf(root->right,leaf);
    }

    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
    
    
        vector<int> leaf1;
        vector<int> leaf2;
        findLeaf(root1,leaf1);
        findLeaf(root2,leaf2);
        return leaf1==leaf2;
    }
};
    }
};

写法2：

/**
 * 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:
    void findLeaf(TreeNode* root, vector<int>& leaf){
    
    
        if(root->left) findLeaf(root->left,leaf);
        if(root->left == nullptr && root->right == nullptr) leaf.push_back(root->val);
        if(root->right) findLeaf(root->right,leaf);
    }

    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
    
    
        vector<int> leaf1;
        vector<int> leaf2;
        findLeaf(root1,leaf1);
        findLeaf(root2,leaf2);
        return leaf1==leaf2;
    }
};