[LeetCode] Symmetric Tree

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.

判断一棵树的对称性，利用递归按照对称树的性质判断即可。

/**
 * 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* root) {
        if (root == nullptr)
            return true;
        return isSymmetricCore(root->left, root->right);
    }
    bool isSymmetricCore(TreeNode* s, TreeNode* t) {
        if (s == nullptr && t == nullptr)
            return true;
        if (s == nullptr || t == nullptr)
            return false;
        if (s->val != t->val)
            return false;
        return isSymmetricCore(s->left, t->right) && isSymmetricCore(s->right, t->left);
    }
};
// 3 ms

 用迭代来表述算法，利用两个queue维护一棵树根节点的左右两棵子树节点。然后判断相对应的节点

/**
 * 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* root) {
        if (root == nullptr)
            return true;
        queue<TreeNode*> q1, q2;
        q1.push(root->left);
        q2.push(root->right);
        while (!q1.empty() && !q2.empty()) {
            TreeNode* node1 = q1.front();
            TreeNode* node2 = q2.front();
            q1.pop();
            q2.pop();
            if ((node1 == nullptr && node2 != nullptr) || (node1 != nullptr && node2 == nullptr))
                return false;
            if (node1 != nullptr && node2 != nullptr) {
                if (node1->val != node2->val)
                    return false;
                q1.push(node1->left);
                q1.push(node1->right);
                q2.push(node2->right);
                q2.push(node2->left);
            }
        }
        return true;
    }
};
// 3 ms