leetcode110(平衡二叉树)

平衡二叉树

给定一个二叉树，判断它是否是高度平衡的二叉树。

本题中，一棵高度平衡二叉树定义为：

一个二叉树每个节点的左右两个子树的高度差的绝对值不超过1。
在这里插入图片描述
返回 false 。

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

*/

class Solution {
public:    
    bool isBalanced(TreeNode *root) {
        if (checkDepth(root) == -1) return false;
        else return true;
    }
    int checkDepth(TreeNode *root) {
        if (!root) return 0;
        int left = checkDepth(root->left);
        if (left == -1) return -1;
        int right = checkDepth(root->right);
        if (right == -1) return -1;
        int diff = abs(left - right);
        if (diff > 1) return -1;
        else return 1 + max(left, right);
    }
};

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

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        return self.depth(root) != -1
    
    def depth(self,root):
        if not root:
            return 0
        left = self.depth(root.left)
        if left == -1:
            return -1
        right = self.depth(root.right)
        if right == -1:
            return -1
        return max(left,right) + 1 if abs(left-right) < 2 else -1
        
*/
  
class Solution:
    def  isBalanced(self,root):
        if not root:
            return True
        return abs(self.depth(root.left)-self.depth(root.right)) <=1 and self.isBalanced(root.left) and self.isBalanced(root.right)
    
    def depth(self,root):
        if not root:
            return 0
        return max(self.depth(root.left),self.depth(root.right)) + 1

来源：CSDN

作者：AI算法工程师YC

链接：https://blog.csdn.net/qq_36134437/article/details/103881767

标签

平衡二叉树