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 }
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 }
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 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsBalanced.java
posted on 2014-12-18 20:54 Yu's Garden 阅读(935) 评论(0) 编辑 收藏 举报