[LeetCode] 101. Symmetric Tree 对称树

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 

Note:
Bonus points if you could solve it both recursively and iteratively.

给一个二叉树判断是否为对称树。

解法1: 非递归，按层遍历，每一层检查一下是否对称。

解法2: 递归，

其中左子树和右子树对称的条件：1）两个节点值相等，或者都为空。2）左节点的左子树和右节点的右子树对称，左节点的右子树和右节点的左子树对称

Java：Recursion

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class Solution {     public boolean isSymmetric(TreeNode root) {       if (root == null)         return true;       return isSymmetric(root.left, root.right);     }       public boolean isSymmetric(TreeNode l, TreeNode r) {       if (l == null && r == null) {         return true;       } else if (r == null || l == null) {         return false;       }         if (l.val != r.val)         return false;         if (!isSymmetric(l.left, r.right))         return false;       if (!isSymmetric(l.right, r.left))         return false;         return true;     } }

Java: Iteration

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class Solution {     public boolean isSymmetric(TreeNode root) {         if(root == null)             return true;         if(root.left == null && root.right == null)             return true;         if(root.left == null || root.right == null)             return false;         LinkedList<TreeNode> q1 = new LinkedList<TreeNode>();         LinkedList<TreeNode> q2 = new LinkedList<TreeNode>();         q1.add(root.left);         q2.add(root.right);         while(!q1.isEmpty() && !q2.isEmpty()){             TreeNode n1 = q1.poll();             TreeNode n2 = q2.poll();               if(n1.val != n2.val)                 return false;             if((n1.left == null && n2.right != null) || (n1.left != null && n2.right == null))                 return false;             if((n1.right == null && n2.left != null) || (n1.right != null && n2.left == null))                 return false;               if(n1.left != null && n2.right != null){                 q1.add(n1.left);                 q2.add(n2.right);             }               if(n1.right != null && n2.left != null){                 q1.add(n1.right);                 q2.add(n2.left);             }                    }         return true;     } }

Python: Recursion

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class Solution:     # @param root, a tree node     # @return a boolean     def isSymmetric(self, root):         if root is None:             return True                   return self.isSymmetricRecu(root.left, root.right)           def isSymmetricRecu(self, left, right):         if left is None and right is None:             return True         if left is None or right is None or left.val != right.val:             return False         return self.isSymmetricRecu(left.left, right.right) and self.isSymmetricRecu(left.right, right.left)

Python: Iteration

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class Solution:     # @param root, a tree node     # @return a boolean     def isSymmetric(self, root):         if root is None:             return True         stack = []         stack.append(root.left)         stack.append(root.right)                   while stack:             p, q = stack.pop(), stack.pop()                           if p is None and q is None:                 continue                           if p is None or q is None or p.val != q.val:                 return False                           stack.append(p.left)             stack.append(q.right)                           stack.append(p.right)             stack.append(q.left)                       return True

C++: Recursion

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/**  * 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) return true;         return isSymmetric(root->left, root->right);     }     bool isSymmetric(TreeNode *left, TreeNode *right) {         if (!left && !right) return true;         if (left && !right || !left && right || left->val != right->val) return false;         return isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);     }       };

C++: Iteration

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class Solution { public:     bool isSymmetric(TreeNode *root) {         if (!root) return true;         queue<TreeNode*> q1, q2;         q1.push(root->left);         q2.push(root->right);                   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);             }         }         return true;     } };

　　　

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