101. Symmetric Tree(树的对称)

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / 
  2   2
      
   3    3

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

方法一：递归
时间复杂度：o（n）             空间复杂度：o（1）

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

方法二：迭代
时间复杂度：o（n）             空间复杂度：o（1）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        Queue<TreeNode> q1=new LinkedList<TreeNode>();
        Queue<TreeNode> q2=new LinkedList<TreeNode>();
        q1.add(root);
        q2.add(root);
        while (!q1.isEmpty()&&!q2.isEmpty()){
            TreeNode t1=q1.poll();
            TreeNode t2=q2.poll();
            if(t1==null&&t2==null) continue;
            if(t1==null||t2==null) return false;
            if(t1.val!=t2.val) return false;
            q1.add(t1.left);
            q1.add(t1.right);
            q2.add(t2.right);
            q2.add(t2.left);
        }
        return true;
    }
}