[LeetCode]Leaf-Similar Trees

递归收集两个数的叶子然后比较即可。时间复杂度O(m + n)，m和n分别为两个树的size，代码如下：

	class Solution {
	public:
	bool leafSimilar(TreeNode* root1, TreeNode* root2) {
	vector<int> res1, res2;
	dfs(root1, res1);
	dfs(root2, res2);
	return res1 == res2;
	}
	private:
	void dfs(TreeNode* node, vector<int>& leaves)
	{
	if (!node)return;
	if (!node->left && !node->right)
	{
	leaves.push_back(node->val);
	return;
	}
	dfs(node->left, leaves);
	dfs(node->right, leaves);
	}
	};

view raw Leaf-Similar Trees.cpp hosted with ❤ by GitHub