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
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { void visitSubTree(TreeNode *root, bool isLeftTree, vector<int> &curVector) { if (!root) return; if (isLeftTree) { visitSubTree(root->left, isLeftTree, curVector); curVector.push_back(root->val); visitSubTree(root->right, isLeftTree, curVector); } else { visitSubTree(root->right, isLeftTree, curVector); curVector.push_back(root->val); visitSubTree(root->left, isLeftTree, curVector); } } public: bool isSymmetric(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if (!root) return true; vector<int> leftVector; vector<int> rightVector; visitSubTree(root->left, true, leftVector); visitSubTree(root->right, false, rightVector); int leftSize = leftVector.size(); int rightSize = rightVector.size(); if (leftSize != rightSize) return false; int idx = 0; while (idx < leftSize) { if (leftVector[idx] == rightVector[idx]) { idx++; } else { return false; } } return true; } };
另外一种解法,转自http://blog.csdn.net/xudli/article/details/8427025
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSymRec(TreeNode *left, TreeNode * right){ if(left == NULL && right==NULL){ return true; } else if(left == NULL || right == NULL) { return false; } return ( left->val == right->val && isSymRec( left->left, right->right) && isSymRec( left->right, right->left) ); //both are not NULL. } bool isSymmetric(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if(root == NULL) return true; return isSymRec( root->left, root->right); } };