标签：情况 blank get ati targe traversal tin 循环 比较
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.
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}"./** * 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) { if(root==nullptr) return true; queue<TreeNode *> Q1; queue<TreeNode *> Q2; Q1.push(root->left); Q2.push(root->right); while( !Q1.empty()) { TreeNode *node1=Q1.front(); TreeNode *node2=Q2.front(); Q1.pop(); Q2.pop(); if(node1==nullptr&&node2==nullptr) continue; if(node1==nullptr||node2==nullptr) return false; if(node1->val !=node2->val) return false; Q1.push(node1->left); Q1.push(node1->right); Q2.push(node2->right); Q2.push(node2->left); } return true; } };
/* //主体的另一种写法
while (!q1.empty() && !q2.empty())
{
TreeNode *node1 = q1.front();
TreeNode *node2 = q2.front();
q1.pop();
q2.pop();
if((node1 && !node2) || (!node1 && node2)) return false;
if (node1)
{
if (node1->val != node2->val) return false;
q1.push(node1->left);
q1.push(node1->right);
q2.push(node2->right);
q2.push(node2->left);
}
}
*/
方法二：使用栈。特别值得注意的是：入栈的顺序，先左左，后右右，交替入栈，这样，出栈时才能保证同时从一层的两头向中间遍历。循环中，注意的条件：左右同时为0时，这种情况重新遍历即可；有一者为0，则说明不对称，返回false；对应的值不相等，也是返回false；到叶节点以后也是重新遍历未访问的结点即可。
1 /** 2 * Definition for binary tree 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 { 14 if(root==NULL||(root->left==NULL&&root->right==NULL)) 15 return true; 16 17 stack<TreeNode *> stk; 18 stk.push(root->left); 19 stk.push(root->right); 20 21 while( !stk.empty()) 22 { 23 TreeNode *rNode=stk.top(); 24 stk.pop(); 25 TreeNode *lNode=stk.top(); 26 stk.pop(); 27 28 if(rNode==NULL&&lNode=NULL) 29 continue; 30 if(rNode==NULL||lNode==NULL) 31 return false; 32 if(rNode->val !=lNode->val) 33 return false; 34 35 //判断是否到叶节点，若是返回遍历其他 36 if(lNode->left==NULL&&lNode->right==NULL 37 &&rNode->left==NULL&&rNode->right==NULL) 38 continue; 39 else 40 { //注意顺序，先左左后右右，交替入栈 41 stk.push(lNode->left); 42 stk.push(rNode->right); 43 stk.push(lNode->right); 44 stk.push(rNode->left); 45 } 46 } 47 return true; 48 } 49 };
方法三：递归
1 /** 2 * Definition for binary tree 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 { 14 if(root==NULL) 15 return true; 16 return subTreeSym(root->left,root->right); 17 } 18 19 bool subTreeSym(TreeNode *lNode,TreeNode *rNode) 20 { 21 if(lNode==NULL&&rNode==NULL) 22 return true; 23 if(lNode==NULL||rNode==NULL) //和下面的那个不能颠倒顺序 24 return false; 25 if(lNode->val !=rNode->val) 26 return false; 27 28 29 return subTreeSym(lNode->left,rNode->right)&&subTreeSym(lNode->right,rNode->left); 30 } 31 32 33 };
[Leetcode] Symmetric tree 对称二叉树
标签：情况 blank get ati targe traversal tin 循环 比较
原文地址：http://www.cnblogs.com/love-yh/p/6995333.html
