110. Balanced Binary Tree
Easy
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
Return false.
扫描二维码关注公众号,回复:
7410559 查看本文章
package leetcode.easy; public class BalancedBinaryTree { @org.junit.Test public void test1() { TreeNode tn11 = new TreeNode(3); TreeNode tn21 = new TreeNode(9); TreeNode tn22 = new TreeNode(20); TreeNode tn33 = new TreeNode(15); TreeNode tn34 = new TreeNode(7); tn11.left = tn21; tn11.right = tn22; tn21.left = null; tn21.right = null; tn22.left = tn33; tn22.right = tn34; tn33.left = null; tn33.right = null; tn34.left = null; tn34.right = null; System.out.print(isBalanced(tn11)); } @org.junit.Test public void test2() { TreeNode tn11 = new TreeNode(1); TreeNode tn21 = new TreeNode(2); TreeNode tn22 = new TreeNode(2); TreeNode tn31 = new TreeNode(3); TreeNode tn32 = new TreeNode(3); TreeNode tn41 = new TreeNode(4); TreeNode tn42 = new TreeNode(4); tn11.left = tn21; tn11.right = tn22; tn21.left = tn31; tn21.right = tn32; tn22.left = null; tn22.right = null; tn31.left = tn41; tn31.right = tn42; tn32.left = null; tn32.right = null; tn41.left = null; tn41.right = null; tn42.left = null; tn42.right = null; System.out.print(isBalanced(tn11)); } public boolean isBalanced(TreeNode root) { if (null == root) { return true; } else { return (Math.abs(maxDepth(root.left) - maxDepth(root.right)) <= 1) && isBalanced(root.left) && isBalanced(root.right); } } private static int maxDepth(TreeNode root) { if (null == root) { return 0; } else { return 1 + Math.max(maxDepth(root.left), maxDepth(root.right)); } } }