115 lines
2.6 KiB
C++
115 lines
2.6 KiB
C++
// Source : https://oj.leetcode.com/problems/symmetric-tree/
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// Author : Hao Chen
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// Date : 2014-06-28
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/**********************************************************************************
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*
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* Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
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*
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* For example, this binary tree is symmetric:
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*
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* 1
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* / \
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* 2 2
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* / \ / \
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* 3 4 4 3
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*
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* But the following is not:
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*
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* 1
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* / \
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* 2 2
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* \ \
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* 3 3
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*
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* Note:
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* Bonus points if you could solve it both recursively and iteratively.
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*
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* confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
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*
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* OJ's Binary Tree Serialization:
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*
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* The serialization of a binary tree follows a level order traversal, where '#' signifies
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* a path terminator where no node exists below.
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*
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* Here's an example:
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*
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* 1
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* / \
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* 2 3
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* /
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* 4
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* \
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* 5
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*
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* The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
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*
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*
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**********************************************************************************/
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/**
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* Definition for binary tree
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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Solution(){
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srand(time(NULL));
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}
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bool isSymmetric(TreeNode *root) {
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if (root == NULL ) return true;
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return isSymmetric(root->left, root->right);
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}
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bool isSymmetric(TreeNode *p, TreeNode *q){
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if (random()%2){
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return isSymmetric1(p, q);
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}
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return isSymmetric2(p, q);
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}
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bool isSymmetric1(TreeNode *p, TreeNode *q){
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if (p==NULL && q==NULL) return true;
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if (p==NULL || q==NULL) return false;
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return (p->val == q->val) &&
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isSymmetric(p->left, q->right) &&
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isSymmetric(p->right, q->left);
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}
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bool isSymmetric2(TreeNode *p, TreeNode *q){
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queue<TreeNode*> q1;
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queue<TreeNode*> q2;
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q1.push(p);
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q2.push(q);
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while(q1.size()>0 && q2.size()>0){
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TreeNode* p1 = q1.front();
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q1.pop();
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TreeNode* p2 = q2.front();
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q2.pop();
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if (p1==NULL && p2==NULL) continue;
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if (p1==NULL || p2==NULL) return false;
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if (p1->val != p2->val) return false;
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q1.push(p1->left);
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q2.push(p2->right);
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q1.push(p1->right);
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q2.push(p2->left);
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}
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return true;
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}
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};
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