题解 | #判断是不是平衡二叉树#
判断是不是平衡二叉树
https://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222
整体思路:左子树是平衡二叉树,右子树是平衡二叉树,左右子树高度差不超过1
- 求高度函数
- 递归判断是不是平衡二叉树
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * }; */ #include <complex> class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param pRoot TreeNode类 * @return bool布尔型 */ bool IsBalanced_Solution(TreeNode* pRoot) { // write code here if (pRoot == nullptr) { return true; } int leftHeight = getHeight(pRoot -> left); int rightHeight = getHeight(pRoot -> right); bool temp = abs(leftHeight - rightHeight) <2?true:false; return IsBalanced_Solution(pRoot -> left) && IsBalanced_Solution(pRoot -> right) && temp; } private: int getHeight(TreeNode* pRoot) { if(pRoot == nullptr) return 0; return max(getHeight(pRoot->left),getHeight(pRoot -> right)) +1; } };