[LeetCode] 872. Leaf-Similar Trees

原题链接： https://leetcode.com/problems/leaf-similar-trees/

1. 题目介绍

Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a leaf value sequence.
将一棵二叉树的全部叶子节点都找出来，并且将它们按照从左到右的顺序，依次排列。

For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).
举个例子，上面这颗二叉树，叶子节点的排列为(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.
如果两个二叉树的叶子节点排列是一样的，那么就认为这两棵树是“叶子相似”的。需要写一个函数判断两个二叉树是不是“叶子相似”。

Note:
Both of the given trees will have between 1 and 100 nodes.
两棵树的节点数都是1到100之间。

2. 解题思路

寻找叶子节点，并且要按照从左至右的顺序排列。一个非常合适的方法就是深度优先搜索。因此需要设计一个深度优先搜索的函数，负责把一棵树的叶子节点排列计算出来。然后再比较这两棵树的排列就可以了。
实现代码

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

        return a.equals(b);
    }
	
	public void treeToList(TreeNode root , ArrayList<Integer> t){
		if (root.left == null && root.right == null) {
			t.add(root.val);
			return;
		}

		if (root.left != null) {
			treeToList(root.left,t);
		}
		if (root.right != null) {
			treeToList(root.right,t);
		}
		return;
	}
}