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
Note:
Bonus points if you could solve it both recursively and iteratively.
宽度遍历
class Solution { public: bool isSymmetric(TreeNode* root) { queue<TreeNode*> left,right; if(!root||(!root->left && !root->right) ) return true; TreeNode* lp;TreeNode* rp; if(root->left) left.push(root->left); if(root->right) right.push(root->right); while(!left.empty()&&!right.empty()) { lp=left.front(); rp=right.front(); if(lp->val!=rp->val) return false; if(lp->left) left.push(lp->left); if(lp->right) left.push(lp->right); if(rp->right) right.push(rp->right); if(rp->left) right.push(rp->left); left.pop(); right.pop(); } if(left.size()!=right.size()) return false; return true; } };此代码存在问题！ 1 / \ 2 2 \ \ 3 3 对于上面的test就无法得出正确答案。正确代码如下：
class Solution { public: bool isSymmetric(TreeNode* root) { if (!root) return true; queue<TreeNode*> check; check.push(root->left); check.push(root->right); while (!check.empty()) { TreeNode* node1 = check.front(); check.pop(); TreeNode* node2 = check.front(); check.pop(); if (!node1 && node2) return false; if (!node2 && node1) return false; if (node1 && node2) { if (node1->val != node2->val) return false; check.push(node1->left); check.push(node2->right); check.push(node1->right); check.push(node2->left); } } return true; } }; 树的遍历的基础。