參考資訊:
https://www.cnblogs.com/grandyang/p/10771842.html
題目:

解答:
/**
* 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 leafSimilar(TreeNode* root1, TreeNode* root2) {
vector<int> v1;
vector<int> v2;
dfs(root1, v1);
dfs(root2, v2);
return v1 == v2;
}
int dfs(TreeNode *n, vector<int> &r) {
if (!n) {
return -1;
}
if (!n->left && !n->right) {
r.push_back(n->val);
return 1;
}
dfs(n->left, r);
dfs(n->right, r);
return 0;
}
};