[LeetCode] 872. Leaf-Similar Trees

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.

Example 1:

Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
Output: true

Example 2:

Input: root1 = [1], root2 = [1]
Output: true

Example 3:

Input: root1 = [1], root2 = [2]
Output: false

Example 4:

Input: root1 = [1,2], root2 = [2,2]
Output: true

Example 5:

Input: root1 = [1,2,3], root2 = [1,3,2]
Output: false

Constraints:

• The number of nodes in each tree will be in the range [1, 200].
• Both of the given trees will have values in the range [0, 200].

叶子相似的树。

请考虑一棵二叉树上所有的叶子，这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。

如果有两棵二叉树的叶值序列是相同，那么我们就认为它们是 叶相似 的。

如果给定的两个根结点分别为 root1 和 root2 的树是叶相似的，则返回 true；否则返回 false 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/leaf-similar-trees
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这是一道easy题，既然比较的是叶子节点，那么我们还是利用DFS去找到每个叶子节点，加入一个list，然后比较两个list是否相同即可。

时间O(n)

空间O(h)

Java实现

/**
 * 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) {
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();
        helper(root1, list1);
        helper(root2, list2);
        return list1.equals(list2);
    }
    
    private List<Integer> helper(TreeNode root, List<Integer> res) {
        if (root == null) {
            return null;
        }
        if (root.left == null && root.right == null) {
            res.add(root.val);
        }
        helper(root.left, res);
        helper(root.right, res);
        return res;
    }
}

LeetCode 题目总结