Yu's Coding Garden : leetcode Question 109: Symmetric Tree

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

From http://rangercyh.blog.51cto.com/1444712/1306485
用了更加明确的方法，用两个栈，左栈保存根节点的左孩子，右栈保存根节点的右孩子，每次从两个栈中取出一个值，然后比较，如果不同则返回false，如果相同则把从左栈中取出的节点的左孩子和右孩子放进左栈；相对的，把右栈中取出的节点的右孩子和左孩子放进右栈。要注意的是左栈取出的节点放孩子的顺序是先左后右，右栈则是先右后左。
From http://gongxuns.blogspot.com/2012/12/leetcodesymmetric-tree.html

    public boolean isSymmetric(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root==null) return true;
        LinkedList<TreeNode> l = new LinkedList<TreeNode>(),
                            r = new LinkedList<TreeNode>();
        l.add(root.left);
        r.add(root.right);
        while(!l.isEmpty() && !r.isEmpty()){
            TreeNode temp1=l.poll(),
                     temp2=r.poll();
            if(temp1==null && temp2!=null || temp1!=null && temp2==null)
                return false;
            if(temp1!=null){
                if(temp1.val!=temp2.val) return false;
                l.add(temp1.left);
                l.add(temp1.right);
                r.add(temp2.right);
                r.add(temp2.left);
            }
        }
        return true;
    }

Recursive Version:
bool isSymmetric(TreeNode *root)
{
if (!root) return true;
return isSymmetric(root->left, root->right);
}
bool isSymmetric(TreeNode *lt, TreeNode *rt)
{
if (!lt && !rt) return true;
if (lt && !rt || !lt && rt || lt->val != rt->val) return false;
return isSymmetric(lt->left, rt->right) &&isSymmetric(lt->right, rt->left);
}
Also refer to: http://rangercyh.blog.51cto.com/1444712/1306485
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