101. Symmetric Tree

题目描述:

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

       1
      / \
       2   2
     / \ / \
   3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

  1
    / \
  2   2
    \   \
    3    3

Note: Bonus points if you could solve it both recursively and iteratively.

题目翻译

给定一颗二叉树,判断它是不是自我镜像,就是以中心为轴镜面对称 比如,这个二叉树 [1,2,2,3,4,4,3] 就是对称的:

       1
      / \
       2   2
     / \ / \
   3  4 4  3

但是如下这个 [1,2,2,null,3,null,3] 就不是:

  1
    / \
  2   2
    \   \
    3    3

注意:如果你可以递归地和迭代地解决它,你就会得到额外的奖励

解题方案

标签:

思路:

解题关键是确定镜面节点对,然后判断该对节点值是否一样即可。通过观察可以发现,只要左子树与右子树镜面对称即可,左子树的左孩子与右子树的右孩子配对,左子树的右孩子与右子树的左孩子配对,因而可以递归求解 也可以迭代求解。

代码:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
   bool isSymmetric(TreeNode *lhs, TreeNode *rhs)
    {
        if (NULL == lhs&&NULL == rhs)return true;
        if (NULL!=lhs&&NULL!=rhs&&lhs->val == rhs->val)
        {
            return isSymmetric(lhs->right, rhs->left) && isSymmetric(lhs->left, rhs->right);
        }
        return false;
    }
    bool isSymmetric(TreeNode* root) {
        if (!root)return true;
        return isSymmetric(root->left, root->right);
    }
};

参考资料

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