LeetCode OJ：Balanced Binary Tree（平衡二叉树）

全栈程序员-用户IM • 2021年9月10日上午6:00 • 未分类

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.

就是去查看一棵树是不是平衡的，一开始对平衡二叉树的理解有错误，所以写错了，看了别人的解答之后更正过来了：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool isBalanced(TreeNode* root) {
13         int dep;
14         checkBalance(root, dep);
15     }
16     bool checkBalance(TreeNode * root, int &dep)
17     {
18         if(root == NULL){
19             dep = 0;
20             return true;
21         }
22         int leftDep, rightDep;
23         bool isLeftBal = checkBalance(root->left, leftDep);
24         bool isRightBal = checkBalance(root->right, rightDep);
25 
26         dep = max(leftDep, rightDep) + 1;
27         return isLeftBal && isRightBal && (abs(leftDep - rightDep) <= 1);
28     }
29 };

pS：感觉这个不应该easy的题目啊想的时候头还挺疼的。。

用java的时候用上面的方法去做总是无法成功，所以换了一种方法，这个一开始没有想到，是看别人写的，代码如下所示：

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 public class Solution {
11     public boolean isBalanced(TreeNode root) {
12         if(root == null)
13             return true; 
14         if(root.left == null && root.right == null)
15             return true;
16         if(Math.abs(getDep(root.left) - getDep(root.right)) > 1)
17             return false;
18         return isBalanced(root.left) && isBalanced(root.right);
19     }
20     
21     
22     public int getDep(TreeNode node){
23         if(node == null)
24             return 0;
25         else
26             return 1 + Math.max(getDep(node.left), getDep(node.right));
27     }
28 }