[LeetCode] 872. Leaf-Similar Trees

[grandyang]

Consider all the leaves of a binary tree.  From left to right order, the values of those leaves form a leaf value sequence.

For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).

Two binary trees are considered  leaf-similar  if their leaf value sequence is the same.

Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.

这道题定义了一种叶相似树，就是说若两棵树的叶结点按照从左向右的顺序取出来排成序列，若两个序列相同，则说明二者是叶结点相似树。其实本质就是按从左到右的顺序打印二叉树的叶结点呗，那么根据这种顺序，我们采用先序遍历遍历比较好，遇到叶结点后直接将叶结点存入数组中，那么对于两个树遍历后就分别得到两个包含叶结点的数组，最后再比较一下这两个数组是否相同即可，参见代码如下：
解法一：

class Solution {
public:
    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
		vector<int> leaf1, leaf2;
		helper(root1, leaf1);
		helper(root2, leaf2);
		return leaf1 == leaf2;
    }
	void helper(TreeNode* node, vector<int>& leaf) {
		if (!node) return;
		if (!node->left && !node->right) {
			leaf.push_back(node->val);
		}
		helper(node->left, leaf);
		helper(node->right, leaf);
	}
};

我们也可以不用数组，而是用两个字符串，那么在每个叶结点值直接要加上一个分隔符，这样才能保证不会错位，最后比较两个字符串是否相等即可，参见代码如下：
解法二：

class Solution {
public:
    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
		string leaf1, leaf2;
		helper(root1, leaf1);
		helper(root2, leaf2);
		return leaf1 == leaf2;
    }
	void helper(TreeNode* node, string& leaf) {
		if (!node) return;
		if (!node->left && !node->right) {
			leaf += to_string(node->val) + "-";
		}
		helper(node->left, leaf);
		helper(node->right, leaf);
	}
};

Github 同步地址:

#872

类似题目：

Binary Tree Preorder Traversal

参考资料：

https://leetcode.com/problems/leaf-similar-trees/

https://leetcode.com/problems/leaf-similar-trees/discuss/152329/C%2B%2BJavaPython-O(logN)-Space

https://leetcode.com/problems/leaf-similar-trees/discuss/152358/Simple-6-lines-Java-StringBuilder-%2B-traverse-with-explanation

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