LeetCode 872. Leaf-Similar Trees-优快云博客

本文链接：https://blog.youkuaiyun.com/github_36324732/article/details/81217971

本文介绍了一种判断两棵二叉树是否叶子相似的方法。通过深度优先搜索（DFS）遍历每棵树的所有叶子节点，并收集它们的值到一个序列中。如果两棵树的叶子值序列相同，则认为这两棵树叶子相似。

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题目链接：https://leetcode.com/problems/leaf-similar-trees/description/

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.

Note:

Both of the given trees will have between 1 and 100 nodes.

题目解析：dfs一下就可以，分别将两棵树的子节点推入vector。

代码如下：0ms Accepted beating 100%

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void dfs(TreeNode* cur, vector<int> &leaves)
    {
        if (cur->left == NULL && cur->right == NULL)
            leaves.push_back(cur->val);
        if (cur->left != NULL)
            dfs(cur->left, leaves);
        if (cur->right != NULL)
            dfs(cur->right, leaves);
    }
    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
        vector<int> t1;
        vector<int> t2;
        dfs(root1, t1);
        dfs(root2, t2);
        if (t1.size() != t2.size())
            return false;
        for (int i = 0; i < t1.size(); i++)
            if (t1[i] != t2[i])
                return false;
        return true;
    }
};