【Leetcode】100. Same Tree

Question：

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally

identical and the nodes have the same value.

My Solution （no recursion）：1ms

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
LinkedList<TreeNode> plist = new LinkedList<TreeNode>();
LinkedList<TreeNode> qlist = new LinkedList<TreeNode>();
public boolean isSameTree(TreeNode p, TreeNode q) {
plist.push(p);
qlist.push(q);
while(plist.size() > 0 && qlist.size() > 0){
TreeNode p1 = plist.pop();
TreeNode q1 = qlist.pop();
if(!compare(p1,q1)){
return false;
}
}
if(plist.size() == qlist.size()){
return true;
}else{
return false;
}

}
public boolean compare(TreeNode p, TreeNode q){
if(p == null && q == null){
return true;
}else if(p != null && q != null){
if(p.val == q.val){
if(compareChild(p.left,q.left)){
if(compareChild(p.right,q.right)){
return true;
}
}
}
}

return false;
}
public boolean compareChild(TreeNode p, TreeNode q){
if(p == null && q == null){
return true;
}else if(p != null && q != null){
if(p.val == q.val){
plist.offer(p);
qlist.offer(q);
return true;
}
}
return false;
}
}

My solution 2 (recursion): 0ms

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if(!compare(p,q)){
return false;
}
return true;
}
public boolean compare(TreeNode p, TreeNode q){
if(p == null && q == null){
return true;
}else if(p != null && q != null){
if(p.val == q.val){
if(compare(p.left,q.left)){
if(compare(p.right,q.right)){
return true;
}
}
}
}

return false;
}

}