Symmetric Tree

1. Question

判断一个二叉树是否是对称的。

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.

OJ‘s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.

Here‘s an example:

   1
  /  2   3
    /
   4
         5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

2. Solution

考虑特殊情况：

• 树为空

采用分治法：

• 左子树顶点和右子树顶点相同
• 左子树顶点的左子树和右子树顶点的右子树对称
• 左子树顶点的右子树和右子树顶点的左子树对称

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    //if two trees are symmetric, return true
    public boolean isSymmetric( TreeNode left, TreeNode right ){
        if( left==null && right==null ) return true;
        if( (left==null && right!=null) || (right==null && left!=null) ) return false;
        return (left.val == right.val) && (isSymmetric( left.left, right.right )) && (isSymmetric( left.right, right.left ));
    }
    public boolean isSymmetric(TreeNode root) {
        if( root == null ) return true;
        return isSymmetric( root.left, root.right );
    }
}