Symmetric Tree

Symmetric Tree

Total Accepted: 84678 Total Submissions: 259420 Difficulty: Easy

 

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.

1.递归版本

/**
 * 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 {
private:
    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->right,root2->left) && isSymmetric(root1->left,root2->right);
    }
public:
    bool isSymmetric(TreeNode* root) {
        return isSymmetric(root,root);
    }
};

 2.迭代版本，巧妙的利用入队列的顺序

/**
 * 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) return true;
        queue<TreeNode*> que;
        que.push(root->left);
        que.push(root->right);
        while(!que.empty()){
            TreeNode* p = que.front();que.pop();
            TreeNode* q = que.front();que.pop();
            if(!p && !q) continue;
            if(!p || !q) return false;
            if(p->val != q->val) return false;
            que.push(p->left);
            que.push(q->right);
            que.push(p->right);
            que.push(q->left);
        }
        return true;
    }
};