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.

package leetcode.easy;

/**
* Definition for a binary tree node. public class TreeNode { int val; TreeNode
* left; TreeNode right; TreeNode(int x) { val = x; } }
*/
public class BalancedBinaryTree {
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));
}
} @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));
}
}

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