标签：style blog color io ar for sp div on
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.
判断二叉树是否是平衡树，比如有两个节点n1, n2，我们需要比较n1的左子节点的值和n2的右子节点的值是否相等，同时还要比较n1的右子节点的值和n2的左子结点的值是否相等，以此类推比较完所有的左右两个节点。我们可以用递归和迭代两种方法来实现，写法不同，但是算法核心都一样。
递归方法 (Recursive Solution)：
/** * 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); } };
迭代方法 (Iterative Solution)：
/** * 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; 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; } };
标签：style blog color io ar for sp div on
原文地址：http://www.cnblogs.com/grandyang/p/4051715.html
