Symmetric Tree

描述

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

  1.     1
  2.    / \
  3.   2   2
  4.  / \ / \
  5. 3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

  1.     1
  2.    / \
  3.   2   2
  4.    \   \
  5.    3    3

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

分析

无

递归版

  1. // Symmetric Tree
  2. // 递归版，时间复杂度O(n)，空间复杂度O(logn)
  3. class Solution {
  4. public:
  5.     bool isSymmetric(TreeNode *root) {
  6.         if (root == nullptr) return true;
  7.         return isSymmetric(root->left, root->right);
  8.     }
  9.     bool isSymmetric(TreeNode *p, TreeNode *q) {
  10.         if (p == nullptr && q == nullptr) return true;   // 终止条件
  11.         if (p == nullptr || q == nullptr) return false;  // 终止条件
  12.         return p->val == q->val      // 三方合并
  13.                 && isSymmetric(p->left, q->right)
  14.                 && isSymmetric(p->right, q->left);
  15.     }
  16. };

迭代版

  1. // Symmetric Tree
  2. // 迭代版，时间复杂度O(n)，空间复杂度O(logn)
  3. class Solution {
  4. public:
  5.     bool isSymmetric (TreeNode* root) {
  6.         if (!root) return true;
  8.         stack<TreeNode*> s;
  9.         s.push(root->left);
  10.         s.push(root->right);
  12.         while (!s.empty ()) {
  13.             auto p = s.top (); s.pop();
  14.             auto q = s.top (); s.pop();
  16.             if (!p && !q) continue;
  17.             if (!p || !q) return false;
  18.             if (p->val != q->val) return false;
  20.             s.push(p->left);
  21.             s.push(q->right);
  23.             s.push(p->right);
  24.             s.push(q->left);
  25.         }
  27.         return true;
  28.     }
  29. };

相关题目

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/binary-tree/traversal/symmetric-tree.html