101. 对称二叉树

/**
 * 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 || root->left == NULL && root->right == NULL) return true;
        return symmetric(root,root);
    }

    //1、两个根节点的值要相等
    //2、左边的左子树和右边的右子树对称
    //3、左边的右子树和右边的左子树对称
    bool symmetric(TreeNode* root1,TreeNode* root2)
    {
        if(root1 == NULL && root2 == NULL) return true;
        if(root1 == NULL || root2 == NULL) return false;
        if(root1->val == root2->val) return symmetric(root1->left,root2->right) && symmetric(root1->right,root2->left);
        else return false;
    }
};

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
//1、左根右遍历
//2、右根左遍历
class Solution 
{
public:
    bool isSymmetric(TreeNode* root) 
    {
        if(root == NULL) return true;
        stack<TreeNode*> s1,s2;
        auto head1 = root->left,head2 = root->right;
        //为什么第二个||不是取&&,因为[1,0]
        while(!s1.empty() || head1 != NULL || !s2.empty() || head2 != NULL)
        {
            if(head1 != NULL && head2 != NULL) 
            {
                s1.push(head1),head1 = head1->left;
                s2.push(head2),head2 = head2->right;
            }
            else if(head1 == NULL && head2 == NULL) 
            {
                head1 = s1.top(),s1.pop();
                int a = head1->val;
                head1 = head1->right;

                head2 = s2.top(),s2.pop();
                int b = head2->val;
                head2 = head2->left;
                if(a != b) return false;
            }
            else return false;
        }
        return true;
    }
};