101. Symmetric Tree

Symmetric Tree

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.

方法一、递归的方法

bool judjeSymmetric(TreeNode* left, TreeNode* right)
{
if(left == NULL && right == NULL)
return true;
else if(left == NULL || right == NULL)
return false;

if(left->val != right->val)
return false;

return judjeSymmetric(left->left,right->right) && judjeSymmetric(left->right,right->left);
}

bool isSymmetric(TreeNode* root) {
if(NULL == root)
return true;
return judjeSymmetric(root->left,root->right);
}

方法二、非递归

bool isSymmetric(TreeNode* root) {
if(NULL == root || (root->left == NULL && root->right == NULL))
return true;

TreeNode* left;
TreeNode* right;
queue<TreeNode*> queLeft;
queue<TreeNode*> queRight;
queLeft.push(root->left);
queRight.push(root->right);
while(!queLeft.empty()&&!queRight.empty())
{
left = queLeft.front();
queLeft.pop();
right = queRight.front();
queRight.pop();
if (NULL == left && NULL == right)
continue;
if (NULL == left || NULL == right)
return false;
if (left->val != right->val)
return false;
queLeft.push(left->left);
queLeft.push(left->right);
queRight.push(right->right);
queRight.push(right->left);
}

return true;
}