【LeetCode】Symmetric Tree (2 solutions)

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.

不管是递归还是非递归，找到关系就好做。

所谓对称，也就是：

1、left对应right

2、left->left对应right->right

3、left->right对应right->left

解法一：递归

/**
 * 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;
        else
            return Helper(root->left, root->right);
    }
    bool Helper(TreeNode* left, TreeNode* right)
    {
        if(!left && !right)
            return true;
        else if(!left && right)
            return false;
        else if(left && !right)
            return false;
        else
        {
            if(left->val != right->val)
                return false;
            else
            {//recursion
                bool partRes1 = Helper(left->left, right->right);
                bool partRes2 = Helper(left->right, right->left);
                return partRes1 && partRes2;
            }
        }
    }
};

解法二：非递归

使用两个队列，对左右子树分别进行层次遍历。

进队时的对应元素比较即可。

/**
 * 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;
        else if(root->left && !root->right)
            return false;   //right is NULL
        else if(!root->left && root->right)
            return false;   //left is NULL
        else if(!root->left && !root->right)
            return true;    //both NULL
        else
        {
            if(root->left->val != root->right->val)
                return false;
            else
            {
                queue<TreeNode*> lq;
                queue<TreeNode*> rq;
                lq.push(root->left);
                rq.push(root->right);
                while(!lq.empty() && !rq.empty())
                {
                    TreeNode* lfront = lq.front();
                    lq.pop();
                    TreeNode* rfront = rq.front();
                    rq.pop();
                    //lfront->left vs. rfront->right
                    if(lfront->left && !rfront->right)
                        return false;   //rfront->right is NULL
                    else if(!lfront->left && rfront->right)
                        return false;   //lfront->left is NULL
                    else if(!lfront->left && !rfront->right)
                        ;   //both NULL
                    else
                    {
                        if(lfront->left->val != rfront->right->val)
                            return false;
                        else
                        {
                            lq.push(lfront->left);
                            rq.push(rfront->right);
                        }
                    }
                    //lfront->right vs. rfront->left
                    if(lfront->right && !rfront->left)
                        return false;   //rfront->left is NULL
                    else if(!lfront->right && rfront->left)
                        return false;   //lfront->right is NULL
                    else if(!lfront->right && !rfront->left)
                        ;   //both NULL
                    else
                    {
                        if(lfront->right->val != rfront->left->val)
                            return false;
                        else
                        {
                            lq.push(lfront->right);
                            rq.push(rfront->left);
                        }
                    }
                }
                return true;
            }
        }
    }
};