LeetCode 101. 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

Follow up: Solve it both recursively and iteratively.

解析：判断一颗二叉树是镜像树，即以根节点左右对称。可以考虑递归和迭代的方法。

递归法：对于两个节点n1和n2，首先判断n1的值是否与n2的值相等，如果不相等在不对称；如果相等，则再判断n1的左子节点和n2的右子节点是否相等，同时n1的右子节点和n2的左子节点是否想相等。代码如下：

/**
 * 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* left, TreeNode* right)
    {
        if(left == NULL && right == NULL)
            return true;
        if(left == NULL or right == NULL)
            return false;
        if(left->val != right->val)
            return false;
        return isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);
    }
    bool isSymmetric(TreeNode* root) {
        if(root == NULL)
            return true;
        return isSymmetric(root->left, root->right);
    }
    
};

迭代法：类似于层次遍历，借助于两个队列，将根节点的左右子树分别放进两个队列中，然后挨个判断是否相等。代码如下：

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

参考：https://www.cnblogs.com/grandyang/p/4051715.html