872. Leaf-Similar Trees

Difficulty:
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Similar Questions:

    Problem

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

      Constraints:

    Solution (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) {
            List<Integer> list1 = new ArrayList<>();
            List<Integer> list2 = new ArrayList<>();
            preOrder(root1, list1);
            preOrder(root2, list2);
            // compare the lists
            if (list1.size() != list2.size()) {
                return false;
            }
            for (int i = 0; i < list1.size(); i++) {
                if (!Objects.equals(list1.get(i), list2.get(i))) {
                    return false;
                }
            }
            return true;
        }
    
        private void preOrder(TreeNode root, List<Integer> list) {
            if (root != null) {
                if (root.left == null && root.right == null) {
                    list.add(root.val);
                }
                preOrder(root.left, list);
                preOrder(root.right, list);
            }
        }
    }
    

    Explain:

    nope.

    Complexity: