(二叉树 DFS 递归) leetcode 101. Symmetric Tree

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.

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symmetric是对称的，此题用DFS会比较简单的，关键是要找出合适的递归方法。

C++代码：

官方题解：https://leetcode.com/problems/symmetric-tree/solution/

/** * 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) { return Recur(root,root); } bool Recur(TreeNode* l,TreeNode* r){ if(l==NULL && r==NULL) return true; if(l==NULL || r==NULL) return false; return (l->val == r->val) && Recur(l->left,r->right) && Recur(l->right,r->left); } };

也可以用迭代，可以用BFS

C++代码：

/** * 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) { queue<TreeNode*> q; if(!root) return true; q.push(root); q.push(root); while(!q.empty()){ auto t1 = q.front(); q.pop(); auto t2 = q.front(); q.pop(); if(!t1 && !t2) continue; if(!t1 || !t2) return false; if(t1->val != t2->val) return false; q.push(t1->left); q.push(t2->right); q.push(t1->right); q.push(t2->left); } return true; } };

 

转载于:https://www.cnblogs.com/Weixu-Liu/p/10738607.html