面试题59：对称的二叉树

/*思路：首先根节点以及其左右子树，左子树的左子树和右子树的右子树相同

* 左子树的右子树和右子树的左子树相同即可，采用递归

* 非递归也可，采用栈或队列存取各级子树根节点

*/

/*public class TreeNode {    int val = 0;    TreeNode left = null;    TreeNode right = null;    public TreeNode(int val) {        this.val = val;    }}*/public class Solution {    boolean isSymmetrical(TreeNode pRoot)    {        return isSymmetrical(pRoot,pRoot);    }        boolean isSymmetrical(TreeNode p1,TreeNode p2){        if(p1==null&&p2==null)            return true;        if(p1==null||p2==null)            return false;        if(p1.val!=p2.val)            return false;         return isSymmetrical(p1.left,p2.right)&&isSymmetrical(p1.right,p2.left);    }}