LeetCode-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.

递归解法：

其中左子树和右子树对称的条件：

1. 两个节点值相等(第二层节点)，或者都为空
2. 左节点的左子树和右节点的右子树对称
3. 左节点的右子树和右节点的左子树对称

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root ==null){
            return true;
        }
        return isSymmetric(root.left, root.right);
        
    }
    public boolean isSymmetric(TreeNode l, TreeNode r){
        if(l == null && r == null){
            return true;
        }
        else if(l == null || r == null){
            return false;
        }
        if(l.val != r.val){
            return false;
        }
        if(!isSymmetric(l.left, r.right)){
            return false;
        }
        if(!isSymmetric(l.right, r.left)){
            return false;
        }
        return true;
    }
}