LeetCode [101] Symmetric Tree

 101. Symmetric Tree

Easy

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

 

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

 

Follow up: Solve it both recursively and iteratively.

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/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: bool isSymmetric(TreeNode* root) { if(!root) return true; return isSym(root->left, root->right); } bool isSym(TreeNode* left, TreeNode* right){ if(!left&&!right) return true; if(!left||!right) return false; if(left->val!=right->val) return false; return isSym(left->left, right->right)&&isSym(left->right, right->left); } };

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/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: bool isSymmetric(TreeNode *root) { if(!root) return true; stack<treenode *> sL; stack<treenode *> sR; sL.push(root->left); sR.push(root->right); while(!sL.empty()){ TreeNode *nL = sL.top(); sL.pop(); TreeNode *nR = sR.top(); sR.pop(); if(!nL&&!nR) continue; if(!(nL&&nR)) return false; if(nL->val!=nR->val) return false; sL.push(nL->right); sL.push(nL->left); sR.push(nR->left); sR.push(nR->right); } return true; } };