[LeetCode] 872. Leaf-Similar Trees 葉結點類似的樹

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.

這道題定義了一種葉類似樹，就是說若兩棵樹的葉結點按照從左向右的順序取出來排成序列，若兩個序列相同，則說明兩者是葉結點類似樹。其實本質就是按從左到右的順序打印二叉樹的葉結點唄，那麼根據這種順序，咱們採用先序遍歷遍歷比較好，遇到葉結點後直接將葉結點存入數組中，那麼對於兩個樹遍歷後就分別獲得兩個包含葉結點的數組，最後再比較一下這兩個數組是否相同便可，參見代碼以下：

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);
    }
};

咱們也能夠不用數組，而是用兩個字符串，那麼在每一個葉結點值直接要加上一個分隔符，這樣才能保證不會錯位，最後比較兩個字符串是否相等便可，參見代碼以下：

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

相似題目：

Binary Tree Preorder Traversal

參考資料：

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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