题目描述
请实现一个函数，用来判断一颗二叉树是不是对称的。注意，如果一个二叉树同此二叉树的镜像是同样的，定义其为对称的。
解:
这道题首先想到的就是按前中后序的一种方法，塞入栈中进行压入弹出的方法
还有一种递归对比左右两树的方法
bool isSymmetricalDFS(TreeNode pRoot) { if(pRoot == null) return true; Stack<TreeNode> s = new Stack<>(); s.push(pRoot.left); s.push(pRoot.right); while(!s.empty()) { TreeNode right = s.pop();//成对取出 TreeNode left = s.pop(); if(left == null && right == null) continue; if(left == null || right == null) return false; if(left.val != right.val) return false; //成对插入 s.push(left.left); s.push(right.right); s.push(left.right); s.push(right.left); } return true; } //改为递归的思路如下 /* struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) { } }; */ class Solution { public: bool isSymmetrical(TreeNode* pRoot) { if( pRoot==nullptr) return true; stack<TreeNode*> tmp; tmp.push(pRoot->left); tmp.push(pRoot->right); return isSymmetrical(tmp); } bool isSymmetrical(stack<TreeNode*>&std_stack) { TreeNode *left= std_stack.top();//成对取出 std_stack.pop(); TreeNode *right= std_stack.top(); std_stack.pop(); if(left==nullptr &&right==nullptr) return true; else if(left==nullptr ||right==nullptr) return false; else if(left->val!=right->val) return false; //两值相等，则塞入堆栈 std_stack.push(left->left); std_stack.push(right->right); std_stack.push(left->right); std_stack.push(right->left); //每次入栈都是两队，递归的话就要调用两次 return isSymmetrical(std_stack)&&isSymmetrical(std_stack); } };
1 /* 2 struct TreeNode { 3 int val; 4 struct TreeNode *left; 5 struct TreeNode *right; 6 TreeNode(int x) : 7 val(x), left(NULL), right(NULL) { 8 } 9 }; 10 */ 11 class Solution { 12 public: 13 bool isSymmetrical(TreeNode* pRoot) 14 { 15 return pRoot==nullptr || isSymmetrical(pRoot->left,pRoot->right); 16 } 17 18 bool isSymmetrical(TreeNode* pRoot1,TreeNode* pRoot2) 19 { 20 if(pRoot1==NULL&&pRoot2==NULL) 21 return true; 22 if(pRoot1==NULL || pRoot2==NULL) 23 return false; 24 if(pRoot1->val!=pRoot2->val) 25 return false; 26 return isSymmetrical(pRoot1->left,pRoot2->right) && isSymmetrical(pRoot1->right,pRoot2->left); 27 28 } 29 30 };