对称树

原题

　　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.

题目大意

　　给定一棵树，判断它是不是对称的。即树的左子树是不是其右子树的镜像。

解题思路

　　使用递归进行求解，先判断左右子结点是否相等，不等就返回false，相等就将左子结点的左子树与右子结果的右子结点进行比较操做，同时将左子结点的左子树与右子结点的左子树进行比较，只有两个同时为真是才返回true，不然返回false。

代码实现

树结点类

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

 

算法实现类

public class Solution {

    public boolean isSymmetric(TreeNode root) {

        if (root == null) {
            return true;
        } else {
            return isSame(root.left, root.right);
        }
    }

    private boolean isSame(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }  if (left != null && right == null || left == null && right != null){
            return false;
        } else {
            return left.val == right.val && isSame(left.left, right.right) && isSame(left.right, right.left);
        }
    }
}