101. Symmetric Tree

1.問題

• 給予一個binary tree, 判斷是否為對稱

2.想法

• 提問

  • 確認題意: root是NULL是true或false?

• function header, parameter

• test input

• 說明想法

  • 將root複製出一份左右對稱的樹後, 比較root 與該數是否是same tree

• 測試計算複雜度

  • recursive的runtime是計算total call number

    • deepCopy

    • swap

    • isSameTree

3.程式碼

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (!root) { 
            return true;
        }
        
        TreeNode* copy = deepCopy(root);
        TreeNode* swapNode = swap(copy);
        return isSameTree(root, swapNode);
    }
private:
    TreeNode* deepCopy(TreeNode* node){
        if (!node) {
            return NULL;
        }
        TreeNode* newNode = new TreeNode(node->val);
        newNode->left = deepCopy(node->left);
        newNode->right = deepCopy(node->right);
        return newNode;
    }
    
    TreeNode* swap(TreeNode* node){
        if (!node) {
            return NULL;
        }
        
        node->left = swap(node->left);
        node->right = swap(node->right);
        TreeNode* tmp = node->left;
        node->left = node->right;
        node->right = tmp;
        return node;
    }
    
    bool isSameTree(TreeNode* p, TreeNode* q) {
        if (!p && !q) {
            return true;
        } 
        
        if (!p || !q) {
            return false;
        } 
        
        if (p->val == q->val) {
            return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
        }
        
        return false;
    }
};