题目

image.png

解法

简单递归。

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void dfs(vector<int>& vec, TreeNode* root) {
        if (!root) {
            return;
        }
        if (!root->left && !root->right) {
            vec.push_back(root->val);
            return;
        }

        dfs(vec, root->left);
        dfs(vec, root->right);
    }

    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
        vector<int> vec1, vec2;
        dfs(vec1, root1);
        dfs(vec2, root2);

        if (vec1.size() != vec2.size()) {
            return false;
        }
        
        for (int i = 0; i < vec1.size(); i++) {
            if (vec1[i] != vec2[i]) {
                return false;
            }
        }

        return true;
    }
};

JAVA

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean leafSimilar(TreeNode root1, TreeNode root2) {
        ArrayList<Integer> list1 = new ArrayList<>();
        ArrayList<Integer> list2 = new ArrayList<>();

        dfs(list1, root1);
        dfs(list2, root2);

        return list1.equals(list2);
    }

    private void dfs(ArrayList<Integer> list, TreeNode root) {
        if (root == null) {
            return;
        }

        if (root.left == null && root.right == null) {
            list.add(root.val);
        }

        dfs(list, root.left);
        dfs(list, root.right);
    }
}