LeetCode101. Symmetric Tree

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   / \  2   2 / \ / \3  4 4  3

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

    1   / \  2   2   \   \   3    3

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

递归

/** * Definition for a binary tree node. * 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 *lt, TreeNode *rt)      {          if (!lt && !rt) return true;          if (lt && !rt || !lt && rt || lt->val != rt->val) return false;          return isSymmetric(lt->left, rt->right) &&isSymmetric(lt->right, rt->left);      }    };

非递归

    bool isSymmetric(TreeNode *root)       {          if (!root || !root->left && !root->right) return true;          TreeNode *t1 = root->left, *t2 = root->right;          if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;          stack<TreeNode *> s1, s2;          s1.push(t1), s2.push(t2);          bool flag = false;          while (!s1.empty() && !s2.empty())          {              if (!flag && (t1->left || t2->right))              {                  s1.push(t1), s2.push(t2);                  t1 = t1->left, t2 = t2->right;                  if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;                  s1.push(t1), s2.push(t2);              }              else if (t1->right || t2->left)              {                  t1 = t1->right, t2 = t2->left;                  if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;                  flag = false;              }              else              {                  t1 = s1.top(), t2 = s2.top();                  s1.pop(), s2.pop();                  flag = true;              }          }          return s1.empty() && s2.empty();      }