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