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Description:
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
Code:
1 bool isSymmetric(TreeNode* root) { 2 if ( root == NULL ) 3 return true; 4 5 stack<TreeNode *>preOrder; 6 stack<TreeNode *>mirrorPreOrder; 7 preOrder.push( root ); 8 mirrorPreOrder.push( root ); 9 10 while( !preOrder.empty() && !mirrorPreOrder.empty() ) 11 { 12 TreeNode * p = preOrder.top(); 13 TreeNode * q = mirrorPreOrder.top(); 14 preOrder.pop(); 15 mirrorPreOrder.pop(); 16 17 if ( p->val != q->val 18 || (p->right ==NULL && q->left) 19 || (p->right && q->left == NULL) 20 || (p->left == NULL && q->right) 21 || (p->left && q->right==NULL)) 22 return false; 23 else 24 { 25 if ( p->right && q->left ) 26 { 27 preOrder.push( p->right ); 28 mirrorPreOrder.push( q->left ); 29 } 30 if (p->left && q->right ) 31 { 32 preOrder.push( p->left ); 33 mirrorPreOrder.push( q->right ); 34 } 35 } 36 } 37 //注意检验是否两棵树都访问完毕 38 if ( preOrder.empty() && mirrorPreOrder.empty() ) 39 return true; 40 else 41 return false; 42 }
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原文地址:http://www.cnblogs.com/happygirl-zjj/p/4589187.html