LeetCode-101 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? > read more on how binary tree is serialized on OJ.

 

题意：判断一颗树是否是镜像对称的。

解法1：使用DFS

public boolean isSymmetric(TreeNode root) {
       if(root == null)
            return true;
        return isSymmetric(root.left, root.right);
    }    

    public boolean isSymmetric(TreeNode left, TreeNode right) {
        if(left == null && right == null)
            return true;
        else if(left == null || right == null || left.val != right.val)
            return false;
        else 
            return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
    }

 

解法2：使用BFS

public boolean isSymmetric(TreeNode root) { 
        if (root == null) { 
            return true; 
        } 
         
        List<TreeNode> result = new ArrayList<TreeNode>(); 
        result.add(root); 
        while (!result.isEmpty()) { 
            // check it is symmetric in same line 
            for (int i=0, j=result.size()-1; i < j; ) { 
                if (result.get(i) == null && result.get(j) == null) { 
                    result.remove(j); 
                    result.remove(i); 
                    j = j-2; 
                } else if (result.get(i) == null || result.get(j) == null || result.get(i).val != result.get(j).val) { 
                    return false; 
                } else { 
                    i++; 
                    j--; 
                } 
            } 
             
            // put next line in the list and remove all current line node 
            int length = result.size(); 
            for (int i=0; i<length; i++) { 
                TreeNode currentNode = result.remove(0); 
                result.add(currentNode.left); 
                result.add(currentNode.right); 
            } 
        } 
        return true; 
    }