Lowest Common Ancestor of a Binary Tree

题目链接

思路：这里来说，我的思路是用递归回溯把每个节点的路径都找出来。代码经过一点小小的手工测试，感觉是对的。但是，leetcode上给我1w个数的测试，然后栈就崩溃了。。在上面看到一个大神的解答，仅仅几行代码，足以看出他对递归的理解不是我能达到的。
我的代码

import java.util.ArrayList;

import javax.sound.midi.Soundbank;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; left=right=null;}
}

public class Solution {


    public static void main(String args[])
    {
        TreeNode root=new TreeNode(1);
        TreeNode temp=new TreeNode(2);
        temp.right=new TreeNode(4);
        temp.right.left=new TreeNode(5);
        root.left=temp;
        root.right=new TreeNode(3);
        Solution so=new Solution();
        TreeNode res=so.lowestCommonAncestor(root, temp.right, temp.right.left);
        System.out.println(res.val);
    }
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        ArrayList<TreeNode> pathsP=new ArrayList<TreeNode>();
        ArrayList<TreeNode> pathsQ=new ArrayList<TreeNode>();


        findPath(root, p, pathsP);
        findPath(root, q, pathsQ);
        if(pathsP.size()==1||pathsQ.size()==1)
        {
            return root;
        }

        int min=pathsP.size()<pathsQ.size()?pathsP.size():pathsQ.size();
        for(int i=0;i<min-1;i++)
        {
            if(pathsP.get(i+1)!=pathsQ.get(i+1))
            {
                return pathsP.get(i);
            }
        }

        if(pathsP.size()<pathsQ.size())
        {
            return pathsP.get(pathsP.size()-1);
        }
        else
        {
            return pathsQ.get(pathsQ.size()-1);
        }
    }

    public boolean findPath(TreeNode root,TreeNode target,ArrayList<TreeNode> paths)
    {

        if(root==null)
        {
            return false;
        }
        paths.add(root);
        if(root.val==target.val)
        {

            return true;
        }
        else
        {
            if(findPath(root.left, target, paths))
            {
                return true;
            }else if(findPath(root.right, target, paths))
            {
                return true;
            }
            else
            {
                paths.remove(root);
                return false;
            }
        }
    }
}

大神的解答

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root)
            return NULL;
        if(root == p || root == q)
            return root;
        // Check if left contains p or q
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        // Check if right contains p or q
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        // if left and right containsp or q the it'sthe LCA if(left && right) return root; return left ? left : right; }