学步园

题目：

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.

confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.

思路:

本题使用了一种很普通的方法，即，先镜像得到镜像后的tree1，然后将原来的tree跟当前tree1进行比较。

如果相同则为真，否则为假。

AC代码：

<pre name="code" class="java">public class Symmetric_Tree {
    private void mirrorSelf(TreeNode root){
        if(root==null)
            return;
        TreeNode left = root.left;
        root.left = root.right;
        root.right = left;
        mirrorSelf(root.left);
        mirrorSelf(root.right);
    }
    private void copy(TreeNode newRoot, TreeNode src){
        if(src!=null){
            TreeNode leftSrc = src.left;
            TreeNode rightSrc = src.right;
            if(leftSrc!=null){
                newRoot.left = new TreeNode(leftSrc.val);
                copy(newRoot.left,src.left);
            }
            if(rightSrc!=null){
                newRoot.right = new TreeNode(rightSrc.val);
                copy(newRoot.right, src.right);
            }
        }
    }
    private boolean equal(TreeNode src, TreeNode src2){
        if(src==null && src2==null)
            return true;
        if(src!=null && src2!=null){
            if(src.val==src2.val){
                return equal(src.left,src2.left) && equal(src.right, src2.right);
            }
        }
        return false;
    }
    public boolean isSymmetric(TreeNode root) {
        if(root==null)
            return true;
        TreeNode src = new TreeNode(root.val);
        copy(src,root);
        mirrorSelf(root);
        return equal(src,root);
    }
}