[LeetCode]101. Symmetric Tree

题目描述：

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / 
  2   2
      
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

思路：

判断一棵树是否对称。bonus：使用递归和迭代两种方法实现
递归：判断根节点是否为空，空则是对称。否则判断左右子节点。
如果左右均为空，则对称。如果左空或右空，则不对称。
如果左子节点的值不等于右子节点，则不对称。
否则比较左子节点的左子节点和右子节点的右子节点，以及左子节点的右子节点和右子节点的左子节点。
依次递归
迭代：
用一个栈或者队列存储树种所有的值，模拟地规定的过程
/*public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}*/

public class Solution101 {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isSymmetric(root.left,root.right);
    }
    public boolean isSymmetric(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;
        return isSymmetric(left.left, right.right) && isSymmetric(left.right,right.left);
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution101 solution101 = new Solution101();
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(2);
        root.left.left = new TreeNode(3);
        root.left.right = new TreeNode(4);
        root.right.left = new TreeNode(4);
        root.right.right = new TreeNode(3);
        if(solution101.isSymmetric(root) == true)
            System.out.println("1");
        else {
            System.out.println("0");
        }
    }

}