Question

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.

Solution

通過觀察，樹的映象是指每個節點的左右子樹互換，先遍歷右子樹，並交換每個節點的左右子樹，然後判斷左右子樹是否相等參考 007-100-判斷兩個二叉樹是否相等 Same Tree

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
void switchNode(struct TreeNode* root){
    if(root != NULL){
        struct
 
 TreeNode* tmp = root->left;
        root->left = root->right;
        root->right = tmp;

        switchNode(root->left);
        switchNode(root->right);
    }
}

bool isSameTree(struct TreeNode* p, struct TreeNode* q) {
    if(p == NULL || q == NULL){
        return (p == q);
    }else
 
{
        if(p->val != q->val){
            return false;
        }else{
            return isSameTree(p->left, q->left)&&isSameTree(p->right, q->right);
        }
    }
}

bool isSymmetric(struct TreeNode* root) {

    if(root != NULL){
        switchNode(root->right);
        return isSameTree(root->left, root->right);
    } 
    return true;
}

時間效率是O(n), 這個解法唯一不好的就是改變了輸入的結構。