Symmetric_Tree.txt

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * 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 isSymmetricHelper(TreeNode *root1, TreeNode *root2){
        if (root1==NULL&&root2==NULL) return true;
        else if(root1==NULL||root2==NULL) return false;
        return (root1->val==root2->val)&&isSymmetricHelper(root1->left,root2->right)&&isSymmetricHelper(root1->right,root2->left);
    }

    bool isSymmetric(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root==NULL) return true;
        return isSymmetricHelper(root->left,root->right);
    }
};

REMARK:
1.Hard one. IDEA: We need a helper function isSymmetricHelper() which takes 2 parameters: root->left and root->right. This helper function try to test if the subtree whose root is root1 is symmetric to the subtree whose root is root2. The process goes: check if root1->val==root2->val and check if subtree root1->left is symmetric to subtree root2->right and check if subtree roo1->right is symmetric to subtree root2->left.
2. isSymmetricHelper() can be pretty error-prone, especially in line34,35,36.


iterator: type
stl::set,map(look up type)

throw exception in destrctor

Inline function

static privavate protect public

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