LeetCode 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.

   1
  /  2   3
    /
   4
         5

思路：递归，判断对称的两个节点是否相同。

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

LeetCode Symmetric Tree

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原文地址：http://blog.csdn.net/u011345136/article/details/45310301