Given two binary trees, write a function to check if they are the same or not.
Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
Example 1:
Input: 1 1 / \ / \ 2 3 2 3 [1,2,3], [1,2,3] Output: true
Example 2:
Input: 1 1 / \ 2 2 [1,2], [1,null,2] Output: false
Example 3:
Input: 1 1 / \ / \ 2 1 1 2 [1,2,1], [1,1,2] Output: false
這個題目利用深度搜索去做,因爲只要發現有一個分支不對稱,那麼就可以判斷這兩個二叉樹不相同
#include <iostream>
using namespace std;
//////鏈表的節點創建,創建一個節點/////////
struct TreeNode {
int val;
TreeNode *lp;
TreeNode *rp;
TreeNode();
TreeNode(int a){val=a;lp=NULL;rp=NULL;}; ////這裏創建一個結構體的構造函數,很方便,初始化的時候直接進行初始化
};
bool isSameTree(TreeNode* p, TreeNode* q) {
if (p == NULL && q == NULL) return true; ////當搜索到最底層的時候,返回true
if (p == NULL || q == NULL) return false;
if (p->val != q->val) return false; /////
return isSameTree(p->lp, q->lp) && isSameTree(p->rp, q->rp); ///這種情況實際上是p和q的val相同的情況,那麼就繼續比較他們的左右子樹,
}
int main() {
TreeNode *a = new TreeNode(1);
TreeNode *b = new TreeNode(2);
TreeNode *c = new TreeNode(3);
TreeNode *d = new TreeNode(4);
TreeNode *e = new TreeNode(5);
TreeNode *f = new TreeNode(6);
TreeNode *a1 = new TreeNode(1);
TreeNode *b1 = new TreeNode(2);
TreeNode *c1= new TreeNode(3);
TreeNode *d1 = new TreeNode(4);
TreeNode *e1 = new TreeNode(5);
TreeNode *f1 = new TreeNode(6);
a->lp = b;
a->rp = c;
b->lp = d;
b->rp = e;
c->lp = f;
a1->lp = b1;
a1->rp = c1;
b1->lp = d1;
// b1->rp = e1;
c1->lp = f1;
isSameTree(a,a1);
return 0;
}
這裏我做的二叉樹是這樣的:
輸入兩個二叉樹:
1 1 / \ / \ 2 3 2 3
/ \ / / /
4 5 6 4 6
通過debug發現,一次搜索的順序是1,2,4,5 然後就結束了,
說明當 return A&&B的時候,只要確認A是false,那麼就沒必要再計算B了,但是如果A是true的話,此時無法確認返回值,則需要繼續計算B,正是因爲語言有這樣的功能,才正真實現了,深度搜索。