【easy】101. Symmetric Tree

判断一棵二叉树是否对称

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
//前序遍历==对称前序遍历
class Solution {
public:bool isSymmetric(TreeNode* root) {return isSymmetric(root,root);}bool isSymmetric(TreeNode* pRoot1,TreeNode* pRoot2){if (pRoot1 == NULL && pRoot2 == NULL)return true;if (pRoot1 == NULL || pRoot2 == NULL)return false;if (pRoot1->val != pRoot2->val)return false;return isSymmetric(pRoot1->left, pRoot2->right) && isSymmetric(pRoot2->left, pRoot1->right);}
};

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
//前序遍历==对称前序遍历
class Solution {
public:bool isSymmetric(TreeNode* root) {return isSymmetric(root,root);}bool isSymmetric(TreeNode* pRoot1,TreeNode* pRoot2){if (pRoot1 == NULL && pRoot2 == NULL)return true;if (pRoot1 == NULL || pRoot2 == NULL)return false;if (pRoot1->val != pRoot2->val)return false;return isSymmetric(pRoot1->left, pRoot2->right) && isSymmetric(pRoot2->left, pRoot1->right);}
};

转载于:https://www.cnblogs.com/sherry-yang/p/8445671.html

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