力扣872 叶子相似的树

   

思路:

直接前序遍历两个树获得叶子节点,然后对vector容器进行比较

   

   

class Solution {

public:

    bool leafSimilar(TreeNode* root1TreeNode* root2) {

        vector<int> r1,r2;

        getLeaf(root1,r1);

        getLeaf(root2,r2);

        return r1==r2;

    }

    void getLeaf(TreeNode* root,vector<int>& ans){

        if(!root->left && !root->right){

            ans.push_back(root->val);

            return;

        }

        if(root->left)

            getLeaf(root->left,ans);

        if(root->right)

            getLeaf(root->right,ans);

    }

};

   

posted on 2022-09-24 15:38  Syukuu  阅读(9)  评论(0)    收藏  举报