leetcode 101 Symmetric Tree
2017-06-23 08:07 tlnshuju 阅读(216) 评论(0) 编辑 收藏 举报
Symmetric Tree Total Accepted: 61440 Total Submissions: 194643 My Submissions
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following is not:
1 / \ 2 2 \ \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}"
means?
> read more on how binary tree is serialized on OJ.
c++ 解决方式:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ #include<queue> using namespace std; typedef pair<TreeNode*,TreeNode*> nodepair; class Solution { public: bool isSymmetricRecursive(TreeNode*a,TreeNode*b){ if(a){ return b && a->val==b->val && isSymmetricRecursive(a->left,b->right) && isSymmetricRecursive(a->right,b->left); } return !b; } bool isSymmetricRecursive(TreeNode*root){ return !root || isSymmetricRecursive(root->left,root->right); } bool isSymmetric(TreeNode *root) { // Level-order BFS. queue<nodepair> q; if(root) q.push(make_pair(root->left,root->right)); while(q.size()){ nodepair p=q.front(); q.pop(); if(p.first){ if(!p.second)return false; if(p.first->val != p.second->val) return false; // the order of children pushed to q is the key to the solution. q.push(make_pair(p.first->left,p.second->right)); q.push(make_pair(p.first->right,p.second->left)); } else if(p.second) return false; } return true; } };
另外一种,非递归解决方式:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSymmetric(TreeNode *root) { TreeNode *left, *right; if (!root) return true; queue<TreeNode*> q1, q2; q1.push(root->left); q2.push(root->right); while (!q1.empty() && !q2.empty()){ left = q1.front(); q1.pop(); right = q2.front(); q2.pop(); if (NULL == left && NULL == right) continue; if (NULL == left || NULL == right) return false; if (left->val != right->val) return false; q1.push(left->left); q1.push(left->right); q2.push(right->right); q2.push(right->left); } return true; } };
/** * 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; stack<TreeNode*> sk; sk.push(root->left); sk.push(root->right); TreeNode* pA, *pB; while(!sk.empty()) { pA = sk.top(); sk.pop(); pB = sk.top(); sk.pop(); if(!pA && !pB) continue; if(!pA || !pB) return false; if(pA->val != pB->val) return false; sk.push(pA->left); sk.push(pB->right); sk.push(pA->right); sk.push(pB->left); } return true; } };
c版本号:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ bool checkNodes(struct TreeNode* a, struct TreeNode* b) { if(a == NULL && b == NULL) { return true; } if(a == NULL || b == NULL) { return false; } if(a->val != b->val) { return false; } return checkNodes(a->left, b->right) && checkNodes(a->right, b->left); } bool isSymmetric(struct TreeNode* root) { if(root == NULL) { return true; } return checkNodes(root->left, root->right); }
递归方案:
bool isSymmetric(TreeNode *root) { if (!root) return true; return helper(root->left, root->right); } bool helper(TreeNode* p, TreeNode* q) { if (!p && !q) { return true; } else if (!p || !q) { return false; } if (p->val != q->val) { return false; } return helper(p->left,q->right) && helper(p->right, q->left); }
python版本号:
class Solution: # @param {TreeNode} root # @return {boolean} def helper(self, a, b): if a is None and b is None: return True if a is None and b is not None: return False if a is not None and b is None: return False if a.val != b.val: return False return self.helper(a.left, b.right) and self.helper(a.right,b.left) def isSymmetric(self, root): if root is None: return True return self.helper(root.left, root.right)
class Solution: # @param {TreeNode} root # @return {boolean} def isSymmetric(self, root): # no tree # is identical if root is None: return True if not self.is_identical(root.left, root.right): return False queue = [] # root is identical # proceed to queue up the next level # (node, depth) if root.left: enqueue(queue, (root.left, 1)) if root.right: enqueue(queue, (root.right, 1)) while queue: same_level = True level = [] while same_level: # still the same level if len(queue) > 0 and (len(level) == 0 or level[-1][1] == queue[0][1]): child = dequeue(queue) level.append(child) # enqueue children now to maintain level order # add to the depth if child[0].left: enqueue(queue, (child[0].left, child[1]+1)) if child[0].right: enqueue(queue, (child[0].right, child[1]+1)) else: same_level = False # symmetrical has to be even if len(level) % 2 != 0: return False while level: # grab the two extreme ends (left_node, _), (right_node, _) = level.pop(0), level.pop() if not self.is_identical(left_node, right_node): return False return True def is_identical(self, left, right): # if any of them is none, they need to be both none if left is None or right is None: return left == right # their value should equal if left.val != right.val: return False # if left has a left, then right needs to have right if left.left: if right.right is None: return False # if left has a right, then right needs to have left if left.right: if right.left is None: return False return True def enqueue(queue, item): queue.append(item) def dequeue(queue): return queue.pop(0)