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文章目录
  1. 1. Balanced Binary Tree
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题

Balanced Binary Tree

题目

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

思路

判断根结点是否平衡，左子树和右子树高度差距小于等于1。然后递归下去判断。

解题

c++版

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/**
 * 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)
            return true;
        int left=height(root->left);
        int right=height(root->right);
        if(left-right<=1 && left-right>=-1)
            return isBalanced(root->left)&&isBalanced(root->right);
        else
            return false;
    }
    
    int height(TreeNode* root){
        if(!root)
            return 0;
        return 1+max(height(root->left),height(root->right));
    }
};

python版

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

class Solution:
    # @param root, a tree node
    # @return a boolean
    def isBalanced(self, root):
        if not root:
            return True
        rightH=self.height(root.right)
        leftH=self.height(root.left)
        if leftH-rightH<=1 and leftH-rightH>=-1:
            return self.isBalanced(root.right) and self.isBalanced(root.left)
        else:
            return False
    
    def height(self,root):
        if not root:
            return 0
        else:
            if not root.right:
                return self.height(root.left)+1
            elif  not root.left:
                return self.height(root.right)+1
            else:
                l = self.height(root.left)  
                r = self.height(root.right)  
                if l<r:
                    return r+1
                else:
                    return l+1

leetcode
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文章目录
  1. 1. Balanced Binary Tree
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题