标签：style blog color 使用 io strong for ar
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.
思路：对二叉树进行遍历即可。递归解法：使用CheckChild( left, right )判断两颗树left和right是否对称，对称条件为：二者都为空，或都不空且left->val==right->val、CheckChild( left->left, right->right ) && CheckChild( left->right, right->left )为真。（即left的左子树、右子树分别与right的右子树、左子树对称）
1 class Solution { 2 public: 3 bool isSymmetric( TreeNode *root ) { 4 if( !root ) { return true; } 5 return CheckChild( root->left, root->right ); 6 } 7 private: 8 bool CheckChild( TreeNode*& left, TreeNode*& right ) { 9 if( !left && !right ) { return true; } 10 if( left && right ) { 11 if( left->val != right->val ) { return false; } 12 return CheckChild( left->left, right->right ) && CheckChild( left->right, right->left ); 13 } 14 return false; 15 } 16 };
使用迭代实现上述递归过程：
1 class Solution { 2 public: 3 bool isSymmetric( TreeNode *root ) { 4 if( !root ) { return true; } 5 queue<TreeNode*> left, right; 6 left.push( root->left ); right.push( root->right ); 7 while( !left.empty() ) { 8 TreeNode *curLeft = left.front(), *curRight = right.front(); 9 left.pop(); right.pop(); 10 if( !curLeft && !curRight ) { continue; } 11 if( curLeft && curRight && curLeft->val == curRight->val ) { 12 left.push( curLeft->left ); left.push( curLeft->right ); 13 right.push( curRight->right ); right.push( curRight->left ); 14 continue; 15 } 16 return false; 17 } 18 return true; 19 } 20 };
标签：style blog color 使用 io strong for ar
原文地址：http://www.cnblogs.com/moderate-fish/p/3917990.html
