[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 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.

思路一：递归思想。时间复杂度O(n)，空间复杂度O(logN)

　　

/**
 * Definition for binary tree
 * 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;
        return isSymmetric(root->left, root->right);
    }
    bool isSymmetric(TreeNode *root1, TreeNode *root2) {
        if (!root1 && !root2) return true;
        if (!root1 || !root2) return false;
        if (root1->val != root2->val) return false;
        
        return isSymmetric(root1->left, root2->right) && isSymmetric(root1->right, root2->left);
    }
};

思路二：迭代。层次遍历的思想。时间复杂度O(n),空间复杂度O(logN）

　　

/**
 * Definition for binary tree
 * 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 *> q;
        q.push(root->left);
        q.push(root->right);
        
        while (!q.empty()) {
            TreeNode *p1 = q.front();
            q.pop();
            TreeNode *p2 = q.front();
            q.pop();
            
            if (!p1 && !p2) continue;
            if (!p1 || !p2) return false;
            if (p1->val != p2->val) return false;
            
            q.push(p1->left);
            q.push(p2->right);
            
            q.push(p1->right);
            q.push(p2->left);
        }
        
        return true;
    }
 
};