LeetCode--Symmetric Tree

Problem:

Given a binary tree, check whether it is a mirror of itself (ie,
symmetric around its center).

For example, this binary tree is symmetric:

              1   
             / \  
            2   2  
           / \ / \ 
          3  4 4  3

But the following is not:

       1   
      / \   
     2   2   
      \   \  
       3   3

Note: Bonus points if you could solve it both recursively and
iteratively. confused what “{1,#,2,3}” means? > read more on how
binary tree is serialized on OJ. Subscribe to see which companies
asked this question

Analysis:
方法1：再次创建一个新的函数，该函数有两个参数，一个是左节点，另一个是右节点，利用这两个记录的节点，分别判断左节点的左孩子？=右节点的右孩子；左节点的右孩子？=右节点的左孩子，将判断的结果返回；
这里其实的对两棵子树同时进行递归判断。
方法2：利用堆栈将节点按照特点的顺序（左节点压左孩子，右节点就得压右孩子；左节点压右孩子，右节点就得压左孩子）压入堆栈，把相应的节点作比较，不相同则返回false，相同的话就continue或者压入堆栈。
Anwser1:
递归：

public class Solution {
    public boolean isSymmetric(TreeNode root) {
    //recursively
        if(root==null) return true;
        else return isSymmetric(root.left,root.right);
    }
    public boolean isSymmetric(TreeNode leftNode,TreeNode rightNode){
        if(leftNode==null && rightNode==null) return true;
        else if (leftNode==null || rightNode==null) return false;
        else {
            if (leftNode.val==rightNode.val) {
                boolean leftSym = isSymmetric(leftNode.left,rightNode.right);
                boolean rightSym = isSymmetric(leftNode.right,rightNode.left);
                return leftSym && rightSym;
            }
            else return false;
        }
    } 
}

Anwser2:
迭代

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null) return true;
        Stack<TreeNode> leftStack = new Stack<TreeNode>();
        Stack<TreeNode> rightStack = new Stack<TreeNode>();
        leftStack.push(root.left);
        rightStack.push(root.right);
        while(leftStack.size()>0 && rightStack.size()>0){
            TreeNode leftNode = leftStack.pop();
            TreeNode rightNode = rightStack.pop();
            if(leftNode==null && rightNode==null) continue;//不能返回true,因为还有其他没有比较的节点。
            else if(leftNode==null || rightNode==null) return false;
            else if(leftNode.val==rightNode.val){
                //压入堆栈的顺序要注意
                leftStack.push(leftNode.left);
                leftStack.push(leftNode.right);
                rightStack.push(rightNode.right);
                rightStack.push(rightNode.left);
            }
            else return false;//考虑的是leftNode.val!=rightNode.val的情况，不能丢
        }
        return true;//考虑的是完全对称的树，最后应该返回true,不能丢
    } 
}