CODE 32. Symmetric Tree

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版权声明：本文为博主原创文章，转载请注明出处：http://blog.jerkybible.com/2013/09/20/2013-09-20-CODE 32 Symmetric Tree/

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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.

OJ’s Binary Tree Serialization:The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.
Here’s an example:

  1
 / \
2   3
   /
  4
   \
    5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

public boolean isSymmetric(TreeNode root) {
	// Start typing your Java solution below
	// DO NOT write main() function
	if (null == root) {
		return true;
	}
	return dfs(root.left, root.right);
}

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

	return (leftResult && rightResult) ? true : false;
}