LeetCode OJ:Symmetric Tree(对称的树)
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
判断一颗树是否对称,首先用递归的方法当然比较容易解决:
1 /** 2 * Definition for a binary tree node. 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 isSymmetric(TreeNode* root) { 13 if(root == NULL) 14 return true; 15 if(!root->left && !root->right) 16 return true; 17 if(root->left && !root->right || !root->left && root->right) 18 return false; 19 return checkSymmetric(root->left, root->right); 20 } 21 22 bool checkSymmetric(TreeNode * left, TreeNode * right) 23 { 24 if(!left && !right) 25 return true; 26 if(!left && right || left && !right) 27 return false; 28 if(left->val != right->val) 29 return false; 30 return checkSymmetric(left->left, right->right) && checkSymmetric(left->right, right->left); 31 } 32 };
题目还要求用非递归的方式来实现,制造两个队列就可以实现了:
1 /** 2 * Definition for a binary tree node. 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 isSymmetric(TreeNode* root) { 13 queue<TreeNode *> q1; 14 queue<TreeNode *> q2; 15 if(root == NULL) return true; 16 if(!root->left && !root->right) return true; 17 if(root->left && !root->right || !root->left && root->right) return false;//val 18 q1.push(root->left); 19 q2.push(root->right); 20 TreeNode * tmpLeft, *tmpRight; 21 while(!q1.empty() && !q2.empty()){ 22 tmpLeft = q1.front(); 23 tmpRight = q2.front(); 24 q1.pop(), q2.pop(); 25 if(!tmpLeft && !tmpRight) 26 continue; 27 if(!tmpLeft && tmpRight || tmpLeft && !tmpRight) 28 return false; 29 if(tmpLeft->val != tmpRight->val) 30 return false; 31 q1.push(tmpLeft->left); 32 q1.push(tmpLeft->right); 33 q2.push(tmpRight->right); 34 q2.push(tmpRight->left); 35 } 36 return q1.empty() && q2.empty(); 37 } 38 };
java的递归版本如下所示:
1 public class Solution { 2 public boolean isSymmetric(TreeNode root) { 3 if(root == null) 4 return true; 5 if(root.left == null && root.right == null) 6 return true; 7 if(root.left != null && root.right != null) 8 return checkSymmetric(root.left, root.right); 9 return false; 10 } 11 12 public boolean checkSymmetric(TreeNode leftNode, TreeNode rightNode){ 13 if(leftNode == null && rightNode == null) 14 return true; 15 if(leftNode != null && rightNode != null){ 16 if(leftNode.val == rightNode.val){ 17 return checkSymmetric(leftNode.left, rightNode.right) && 18 checkSymmetric(leftNode.right, rightNode.left); 19 } 20 else 21 return false; 22 } 23 return false; 24 } 25 }
下面是迭代版本,和上面的基本上一样:
public class Solution { public boolean isSymmetric(TreeNode root) { Queue<TreeNode> q1 = new LinkedList<TreeNode>(); Queue<TreeNode> q2 = new LinkedList<TreeNode>(); TreeNode p1, p2; if(root == null) return true; q1.add(root.left); q2.add(root.right); while(!q1.isEmpty() && !q2.isEmpty()){ p1 = q1.poll(); p2 = q2.poll(); if(p1 == null && p2 == null) continue; if(p1 != null && p2 != null){ if(p1.val == p2.val){ q1.add(p1.left); q1.add(p1.right); q2.add(p2.right); q2.add(p2.left); }else return false; }else return false; } return true; } }