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【leetcode】Symmetric Tree

2024-11-18 04:31:39   202人阅读

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

对称的递归表达式：

testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);

leftNode->left->val==rightNode->right->val

leftNode->right->val==rightNode->left->val

 

 1 class Solution { 2  3 public: 4  5     bool isSymmetric(TreeNode *root) { 6  7         if(root==NULL) 8  9         {10 11             return true;12 13         }14 15         return testMirror(root->left,root->right);16 17        18 19     }20 21    22 23     bool testMirror(TreeNode *leftNode,TreeNode *rightNode)24 25     {26 27        28 29         if(leftNode==NULL&&rightNode==NULL)30 31             return true;32 33         if(leftNode!=NULL&&rightNode==NULL||leftNode==NULL&&rightNode!=NULL)34 35             return false;36 37         if(leftNode->val!=rightNode->val)38 39             return false;40 41            42 43         return testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);44 45        46 47        48 49     }50 51 };

 

 

 

 

 

【leetcode】Symmetric Tree


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