leetcode——110. 平衡二叉树

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        if not root:
            return True
        def helper(node):
            if not node:
                return 0
            left=helper(node.left)
            right=helper(node.right)
            return max(left,right)+1
        return abs(helper(root.left)-helper(root.right))<=1 and self.isBalanced(root.left) and self.isBalanced(root.right)

执行用时 :76 ms, 在所有 python3 提交中击败了60.84%的用户

内存消耗 :19.6 MB, 在所有 python3 提交中击败了5.15%的用户

 

——2019.11.15

 

 

简单题还是可以掌握的，逻辑也简单

public boolean isBalanced(TreeNode root) {
        if(root == null){
            return true;
        }
        return Math.abs(height(root.left) - height(root.right))<2 && isBalanced(root.left) && isBalanced(root.right);
    }

    //如何获取一棵树的高度
    private int height(TreeNode root) {
        if(root == null){
            return 0;
        }else{
            return Math.max(height(root.left),height(root.right))+1;
        }
    }

——2020.7.1