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




Note:
Bonus points if you could solve it both recursively and iteratively.


confused what "{1,#,2,3}" means?
Solution1: recursion

Run time complexity is O(n), space complexity is stack space O(logn).

/**
 * 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 helper(root.left, root.right);
    }
    
    public boolean helper(TreeNode left, TreeNode right) {
        if(left == null && right == null)
            return true;
        if(left == null || right == null)
            return false;
        if(left.val != right.val)
            return false;
        return helper(left.right, right.left) && helper(left.left, right.right);
    }
}

Solution2: iterative
用两个queue来保存左右节点，一层一层遍历。

Run time complexity is O(n), space complexity is O(n).

/**
 * 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;
        if(root.left == null && root.right == null)
            return true;
        if(root.left == null || root.right == null)
            return false;
        
        LinkedList queue1 = new LinkedList();
        LinkedList queue2 = new LinkedList();
        queue1.add(root.left);
        queue2.add(root.right);
        while(!queue1.isEmpty() && !queue2.isEmpty()) {
            TreeNode curL = queue1.poll();
            TreeNode curR = queue2.poll();
            if(curL.left == null && curR.right != null || curL.left != null && curR.right == null)
                return false;
            if(curL.right == null && curR.left != null || curL.right != null && curR.left == null)
                return false;
            if(curL.val != curR.val)
                return false;
            
            if(curL.left != null && curR.right != null) {
                queue1.add(curL.left);
                queue2.add(curR.right);
            }
            if(curL.right != null && curR.left != null) {
                queue1.add(curL.right);
                queue2.add(curR.left);
            }
        }
        return true;
    }
}