给定一棵二叉树,判断琪是否是自身的镜像(即:是否对称)两种方法递归与迭代

判断二叉树是否对称

http://www.nowcoder.com/questionTerminal/1b0b7f371eae4204bc4a7570c84c2de1

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param root TreeNode类 
     * @return bool布尔型
     */
      public boolean isSymmetric(TreeNode root) {
        return checkMetric(root, root);
    }

    private boolean checkMetric(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null)
            return true;
        if ((root1 != null && root2 == null) || root1 == null && root2 != null)
            return false;
        return (root1.val == root2.val && checkMetric(root1.left, root2.right) && checkMetric(root1.right, root2.left));
    }
}
// 非递归,利用层次遍历,但是要注意左右入队次序;
        if (root == null)
            return true;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if (root.left == null || root.right == null) {
            return root.left == root.right;
        }
        queue.add(root.left);
        queue.add(root.right);
        while (queue.size() > 0) {
            TreeNode lefTreeNode = queue.poll();
            TreeNode rigtheTreeNode = queue.poll();
            if ((lefTreeNode == null && rigtheTreeNode != null) || (lefTreeNode != null && rigtheTreeNode == null)) {
                return false;
            } else if (lefTreeNode == null && rigtheTreeNode == null) {
                continue;
            } else if (lefTreeNode.val != rigtheTreeNode.val) {
                return false;
            }
            queue.add(lefTreeNode.left);
            queue.add(rigtheTreeNode.right);
            queue.add(lefTreeNode.right);
            queue.add(rigtheTreeNode.left);
        }
        return true;
    }
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