平衡二叉树检查 牛客网 程序员面试金典 C++ Python

平衡二叉树检查 牛客网 程序员面试金典 C++ Python

• 题目描述

• 实现一个函数，检查二叉树是否平衡，平衡的定义如下，对于树中的任意一个结点，其两颗子树的高度差不超过1。

• 给定指向树根结点的指针TreeNode* root，请返回一个bool，代表这棵树是否平衡。

C++

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/

class Balance {
public:
//rum:3ms memory:408k
    bool isBalance(TreeNode* root) {
        if (NULL == root) return true;
        if (NULL == root->left && NULL == root->right) return true;
        if (NULL != root->left && NULL == root->right)
            if(getTreeHeight(root->left) > 1) return false;
        if (NULL == root->left && NULL != root->right)
            if(getTreeHeight(root->right) >1) return false;
        return isBalance(root->left) && isBalance(root->right);
    }
    int getTreeHeight(TreeNode* root){
        if (NULL == root) return 0;
        return max(getTreeHeight(root->left),getTreeHeight(root->right))+ 1;
    }
};

Python

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Balance:
    #run:27ms memory:5736k
    def isBalance(self, root):
        if None == root: return True
        if None == root.left and None == root.right: return True
        if None != root.left and None == root.right:
            if self.getTreeHeight(root.left) > 1: return False
        if None == root.left and None != root.right:
            if self.getTreeHeight(root.right) > 1:return False
        return self.isBalance(root.left) and self.isBalance(root.right) + 1

    def getTreeHeight(self,root):
        if None == root: return 0
        return max(self.getTreeHeight(root.left),self.getTreeHeight(root.right)) + 1