LeetCode刷题Easy篇 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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

我的尝试

用递归的思路，前面写过一个就二叉树最大深度的算法，利用递归计算出左子树和右子树的最大深度，判断abs绝对值是否小于等于1.代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public int maxDepth(TreeNode node){
        if(node==null) return 0;
        return Math.max(maxDepth(node.left),maxDepth(node.right))+1;
        
    }
    
    public boolean isBalanced(TreeNode root) {
        if(root==null) return true;
        return Math.abs(maxDepth(root.left)-maxDepth(root.right))<=1;
    }
}

但是有问题，我理解的情况有偏差，如果下面的情况不正确，因为节点2的左子树是3，4但是右子树为空。

我这个算法，比较的是左侧的最大深度和右侧的最大深度差值，如果是左侧的两个深度比较呢？显然没有覆盖这个情况。

下面进行修改：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public int maxDep(TreeNode node){
        if(node==null) return 0;
        return 1+Math.max(maxDep(node.left),maxDep(node.right));
        
    }
    
    public boolean isBalanced(TreeNode root) {
        if(root==null) return true;
        int diff=Math.abs(maxDep(root.left)-maxDep(root.right));
        if(diff>1){
            return false;
        }
        return isBalanced(root.left)&& isBalanced(root.right);
    }
}