problem
Same Treecode
回歸的方法
/**
* 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 isSameTree(TreeNode* p, TreeNode* q) {
if(p==NULL || q==NULL) return(p==q);
return ( (p->val==q->val) && isSameTree(p->left, q->left) && isSameTree(p->right, q->right) );
}
};
or just one line code
class Solution {
public:
bool isSameTree(TreeNode* p, TreeNode* q) {
return (p==NULL||q==NULL) ? (p==q) : ((p->val==q->val) && isSameTree(p->left, q->left) && isSameTree(p->right, q->right));
}
};
這道題還有非遞歸的解法,因為二叉樹的四種周遊(層序,先序,中序,後序)均有各自的疊代和遞歸的寫法,這裡我們先來看先序的疊代寫法,相當于同時周遊兩個數,然後每個節點都進行比較.