LeetCode -- Symmetric Tree

Question:

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.

 

Analysis:

问题描述：给出一棵二叉树，判断它是否是自己的镜像树。即围绕着中心节点对称。

思路一：遍历整棵二叉树，然后判断每层的节点是否是对称的。

思路二：递归判断。首先判断该节点的左右节点是否对称，然后保存左右节点，依次递归判断左节点的左节点与右节点的右节点是否对称，以及左节点的右节点与右节点的左节点是否对称。

 

Answer：

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

 

转载于:https://www.cnblogs.com/little-YTMM/p/4854956.html