【LeetCode题目记录-11】判断二叉树是否是镜像的（对称的）

Symmetric Tree

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.

【分析1-原创】采用递归方案，查看了参考才搞定。

参考：https://oj.leetcode.com/discuss/456/recusive-solution-symmetric-optimal-solution-inordertraversal

public static boolean isSymmetric(TreeNode root) {
         if(root==null) return true;
         return checkIsSymmetric(root.left,root.right);
    }
    private static boolean checkIsSymmetric(TreeNode leftNode,TreeNode rightNode) {
       if(leftNode==null&&rightNode==null) return true;
       if((leftNode!=null&&rightNode==null)||(leftNode==null&&rightNode!=null)) return false;
       if(leftNode.val!=rightNode.val) return false;
       return checkIsSymmetric(leftNode.left,rightNode.right)&&checkIsSymmetric(leftNode.right,rightNode.left);
} 

【分析2-非原创】用stack对左右进行进stack，然后每一层进行比较。

参考：https://oj.leetcode.com/discuss/456/recusive-solution-symmetric-optimal-solution-inordertraversal

 public boolean isSymmetric(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (root == null)
            return true;
        Stack<TreeNode> s1 = new Stack<TreeNode>();
        Stack<TreeNode> s2 = new Stack<TreeNode>();
        s1.push(root.left);
        s2.push(root.right);
        while (!s1.empty() && !s2.empty()) {
            TreeNode n1 = s1.pop();
            TreeNode n2 = s2.pop();
            if (n1 == null && n2 == null)
                continue;
            if (n1 == null || n2 == null)
                return false;
            if (n1.val != n2.val)
                return false;
            s1.push(n1.left);
            s2.push(n2.right);
            s1.push(n1.right);
            s2.push(n2.left);
        }
        return true;
    }