[LeetCode] Same Tree

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

思路一：前序递归的思想。时间复杂度O(n),空间复杂度O(logN)

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool isSameTree(TreeNode *p, TreeNode *q) {
13         if (!p && !q) return true;
14         if (!p || !q ) return false;
15
16         return (p->val == q->val) && isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
17     }
18 };

思路二：层次遍历的思想。时间复杂度O(n), 空间复杂度O(logN)

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool isSameTree(TreeNode *p, TreeNode *q) {
13         queue<TreeNode *> s;
14         s.push(p);
15         s.push(q);
16
17         while (!s.empty()) {
18             p = s.front();
19             s.pop();
20             q = s.front();
21             s.pop();
22             if (!p && !q) continue;
23             if (!p || !q) return false;
24             if (p->val != q->val) return false;
25
26             s.push(p->left);
27             s.push(q->left);
28             s.push(p->right);
29             s.push(q->right);
30         }
31
32         return true;
33     }
34 };

转载于:https://www.cnblogs.com/vincently/p/4232467.html

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