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leetcode 872 叶子相似的树

leetcode 872 叶子相似的树

作者: 奥利奥蘸墨水 | 来源:发表于2021-05-11 00:00 被阅读0次

    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);
        }
    }
    

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