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/**

* 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.

*/

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

public class BalancedBinaryTree110 {

public boolean isBalanced(TreeNode root) {

if (root == null) {

return true;

}

int depthL = depth(root.left);

int depthR = depth(root.right);

return Math.abs(depthL - depthR) <= 1 && isBalanced(root.left) && isBalanced(root.right);

}

private int depth(TreeNode node) {

if (node == null) {

return 0;

}

int depthL = depth(node.left);

int depthR = depth(node.right);

return Math.max(depthL, depthR) + 1;

}

/**

* https://discuss.leetcode.com/topic/11007/java-solution-based-on-height-check-left-and-right-node-in-every-recursion-to-avoid-further-useless-search

*/

public boolean isBalanced2(TreeNode root) {

if (root == null) {

return true;

}

return height(root) != -1;

}

public int height(TreeNode node){

if (node == null) {

return 0;

}

int lH = height(node.left);

if (lH == -1) {

return -1;

}

int rH = height(node.right);

if (rH == -1) {

return -1;

}

if (lH-rH < -1 || lH-rH > 1) {

return -1;

}

return Math.max(lH,rH) + 1;

}

public static void main(String[] args) {

BalancedBinaryTree110 bbt = new BalancedBinaryTree110();

TreeNode root1 = new TreeNode(2);

root1.left = new TreeNode(1);

root1.right = new TreeNode(3);

TreeNode root3 = new TreeNode(2);

root3.left = new TreeNode(1);

TreeNode root2 = new TreeNode(2);

root2.left = root3;

TreeNode n1 = new TreeNode(1);

TreeNode n2 = new TreeNode(2);

TreeNode n3 = new TreeNode(3);

TreeNode n4 = new TreeNode(4);

TreeNode n5 = new TreeNode(5);

TreeNode n6 = new TreeNode(6);

TreeNode n7 = new TreeNode(7);

TreeNode n8 = new TreeNode(8);

n4.left = n7;

n2.left = n4;

n2.right = n5;

n1.left = n2;

n6.right = n8;

n3.right = n6;

n1.right = n3;

System.out.println("-------- 1");

System.out.println(bbt.isBalanced(root1));

System.out.println("-------- 2");

System.out.println(bbt.isBalanced(root2));

System.out.println("-------- 3");

System.out.println(bbt.isBalanced(root3));

System.out.println("-------- 4");

System.out.println(bbt.isBalanced(n1));

}

}