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  • 【leetcode】Symmetric Tree

    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 };
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  • 原文地址:https://www.cnblogs.com/reachteam/p/4251662.html
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