【leetcode】Balanced Binary Tree

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.

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题解：

一种方法是写一个递归求高度的函数，然后再写一个递归函数判断树是否是平衡的。

代码如下：

 1 public class Solution {

 2     private int height(TreeNode root){

 3         if(root == null)

 4             return 0;

 5         int left = height(root.left);

 6         int right = height(root.right);

 7         

 8         return Math.max(left, right)+1;

 9     }

10     public boolean isBalanced(TreeNode root) {

11         if(root == null)

12             return true;

13         int left = height(root.left);

14         int right = height(root.right);

15         

16         if(Math.abs(left-right)>1)

17             return false;

18         return isBalanced(root.left) & isBalanced(root.right);

19     }

20 }

这种方法耗时428ms。

第二种方法在递归求树的高度的过程中顺便判断树是否平衡，如果在某个节点处，该节点的左子树和右子树高度只差大于1，或者该树的左子树或者又子树不平衡，那么返回该树的高度为-1；否则返回该树的高度。

代码如下：

 1 public class Solution {

 2     private int height(TreeNode root){

 3         if(root == null)

 4             return 0;

 5         int left = height(root.left);

 6         int right = height(root.right);

 7         

 8         if(left == -1 || right == -1 || Math.abs(left - right) > 1)

 9             return -1;

10         return Math.max(left, right)+1;

11     }

12     public boolean isBalanced(TreeNode root) {

13         return height(root) != -1;

14     }

15 }

这种方法耗时464ms。

第二遍刷leetcode时候的java代码也放上来，耗时250ms。

 1 public class Solution {

 2     public boolean isBalanced(TreeNode root) {

 3         height(root);

 4         return isBlanced;

 5     }

 6     boolean isBlanced = true;

 7     private int height(TreeNode root){

 8         if(root == null)

 9             return 0;

10         int leftheight = height(root.left);

11         int rightheight = height(root.right);

12         if(Math.abs(leftheight - rightheight) > 1){

13             isBlanced = false;

14         }

15         return Math.max(leftheight, rightheight)+1;

16     }

17 }