【Leetcode】Symmetric Tree (Tree Judge)

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.

helper作用用于判断每个根节点的左右子节点是否对称。

先判断同向，即左左右右，然后再判断异向，即左右左右。

压栈思路：先左子树左节点和右子树右节点全入，然后左子树左节点出右节点入，右子树右节点出左节点入。在压栈的过程中先后比较左右节点是否为空，并且比较值是否相等。

特殊情况：空数为对称树。子树都为空树，则相等，只有一个为空树，则一定不相等

    public boolean isSymmetric(TreeNode root) {        if(root==null)            return true;        return helper(root.left, root.right);    }        public boolean helper(TreeNode left, TreeNode right) {        if(left==null&&right==null)            return true;        if(left==null||right==null)            return false;        if(left.val!=right.val)            return false;        boolean llrr = helper(left.left, right.right);        boolean lrrl = helper(left.right, right.left);        return llrr&&lrrl;    }