55-02 判断平衡二叉树

题目

输入一课二叉树的根结点，判断该树是不是平衡二叉树。如果二叉树中任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

LeetCode

C++ 题解

采用递归的方法：

/**
 * 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 isBalanced(TreeNode* root) {
        if(root == nullptr)
            return true;
        
        if(abs(getDepth(root->left) - getDepth(root->right)) > 1)
           return false;
           
        return isBalanced(root->left) && isBalanced(root->right);
        
    }
    
    
    int getDepth(TreeNode* root)
    {
        if (root==nullptr)
            return 0;
        
        return max(getDepth(root->left),getDepth(root->right)) +1;        
    }
    
};

python 题解

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        
        if root == None:
            return True       
        
        left = self.GetDepth(root.left)
        right = self.GetDepth(root.right)
        
        if(abs(left - right) > 1):
            return False         
        
        return self.isBalanced(root.left) and self.isBalanced(root.right)
    
            
    def GetDepth(self,proot):
        if proot == None:
            return 0          
        
        left = self.GetDepth(proot.left)
        right = self.GetDepth(proot.right)

        return max(left,right) + 1