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

参考 http://www.cnblogs.com/feiling/p/3251061.html & http://fisherlei.blogspot.com/2013/01/leetcode-symmetric-tree.html

非递归解法：按层遍历，每一层检查一下是否对称。

/**
 * Definition for binary tree
 * 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;
        }
        
        LinkedList<TreeNode> l = new LinkedList<TreeNode>(),
        LinkedList<TreeNode> r = new LinkedList<TreeNode>();
         
         l.add(root.left);
         r.add(root.right);
         
         while(!l.isEmpty() && !r.isEmpty()) {
             TreeNode t1 = l.poll(); TreeNode t2 = r.poll();
             
             if((t1 == null && t2 != null)||(t2 == null && t1 != null)){
                 return false;
             }
             
             if( t1 != null){
                 if(t1.val != t2.val){
                     return false;
                 }
                 
                 l.add(t1.left);
                 l.add(t1.right);
                 r.add(t2.right);
                 r.add(t2.left);
             }
             
         }
         
         return true;
    }
}

递归解法：

/**
 * Definition for binary tree
 * 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 checkSymmetric(root.left, root.right);
    }
    
    private boolean checkSymmetric(TreeNode left, TreeNode right){
        if(left == null && right == null)
            return true;
        
        if((left == null && right != null) || (left != null && right == null))
            return false;
        
        if(left.val != right.val)
            return false;
        
        return checkSymmetric(left.left, right.right) && checkSymmetric(left.right, right.left);
    }
}

其中左子树和右子树对称的条件：

1. 两个节点值相等，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 binary tree
    * 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;
           else
               return check(root.left, root.right);
       }
       
       private boolean check(TreeNode left, TreeNode right){
           if(left == null) return right == null;
           if(right == null) return left == null;
           
           if(left.val != right.val)
               return false;
           else
               return check(left.left, right.right) && check(left.right, right.left);
       }
   }