[LeetCode] 872. Leaf-Similar Trees 叶结点类似的树

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类似树，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，则说明两者是叶结点类似树。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，咱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获得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便可，参见代码以下：

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);
    }
};

咱们也能够不用数组，而是用两个字符串，那么在每一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便可，参见代码以下：

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:

https://github.com/grandyang/leetcode/issues/872

相似题目：

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

LeetCode All in One 题目讲解汇总(持续更新中...)