LeetCode: Balanced Binary Tree 解题报告

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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SOLUTION 1:

使用inner Class来解决

 1 // Solution 1:
 2     public boolean isBalanced1(TreeNode root) {
 3         return dfs(root).isBalanced;
 4     }
 5     
 6     // bug 1: inner class is like:  "public class ReturnType {", no ()
 7     public class ReturnType {
 8         boolean isBalanced;
 9         int depth;
10         
11         ReturnType(int depth, boolean isBalanced) {
12             this.depth = depth;
13             this.isBalanced = isBalanced;
14         }
15     }
16     
17     public ReturnType dfs(TreeNode root) {
18         ReturnType ret = new ReturnType(0, true);
19         
20         if (root == null) {
21             return ret;
22         }
23         
24         ReturnType left = dfs(root.left);
25         ReturnType right = dfs(root.right);
26         
27         ret.isBalanced = left.isBalanced 
28                          && right.isBalanced 
29                          && Math.abs(left.depth - right.depth) <= 1;
30                          
31         // bug 2: remember to add 1( the root depth )                 
32         ret.depth = Math.max(left.depth, right.depth) + 1;
33         
34         return ret;
35     }
View Code

SOLUTION 2:

将 get depth函数提出

 1 // Solution 2:
 2     public boolean isBalanced(TreeNode root) {
 3         if (root == null) {
 4             return true;
 5         }        
 6         
 7         return isBalanced(root.left) && isBalanced(root.right)
 8             && Math.abs(getDepth(root.left) - getDepth(root.right)) <= 1;
 9     }
10     
11     public int getDepth(TreeNode root) {
12         if (root == null) {
13             return 0;
14         }
15         
16         return Math.max(getDepth(root.left), getDepth(root.right)) + 1;
17     }
View Code

SOLUTION 3:

leetcode又加强了数据，solution 2对于一条单链过不了了。所以主页君加了一点优化，当检测到某个子节点为null时，求另一个子树的depth时，及时退出，这

样就不会产生getdepth太深的问题：

 1 // Solution 2:
 2     public boolean isBalanced(TreeNode root) {
 3         if (root == null) {
 4             return true;
 5         }
 6         
 7         boolean cut = false;
 8         if (root.right == null || root.left == null) {
 9             cut = true;
10         }
11         
12         return isBalanced(root.left) && isBalanced(root.right)
13             && Math.abs(getDepth(root.left, cut) - getDepth(root.right, cut)) <= 1;
14     }
15     
16     public int getDepth(TreeNode root, boolean cut) {
17         if (root == null) {
18             return -1;
19         }
20         
21         if (cut && (root.left != null || root.right != null)) {
22             // if another tree is not deep, just cut and return fast.
23             // Improve the performance.
24             return 2;
25         }
26         
27         return 1 + Math.max(getDepth(root.left, false), getDepth(root.right, false));
28     }
View Code

 

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsBalanced.java

posted on 2014-12-18 20:54  Yu's Garden  阅读(934)  评论(0)  编辑  收藏  举报

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