# 110. Balanced Binary Tree

## Description

Given a binary tree, determine if it is height-balanced.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: true

Example 2:

Input: root = [1,2,2,3,3,null,null,4,4]
Output: false

Example 3:

Input: root = []
Output: true

Constraints:

• The number of nodes in the tree is in the range [0, 5000].
• -104 <= Node.val <= 104

## Solutions

Solution 1: Bottom-Up Recursion

We define a function $height(root)$ to calculate the height of a binary tree, with the following logic:

• If the binary tree $root$ is null, return $0$.
• Otherwise, recursively calculate the heights of the left and right subtrees, denoted as $l$ and $r$ respectively. If either $l$ or $r$ is $-1$, or the absolute difference between $l$ and $r$ is greater than $1$, then return $-1$. Otherwise, return $max(l, r) + 1$.

Therefore, if the function $height(root)$ returns $-1$, it means the binary tree $root$ is not balanced. Otherwise, it is balanced.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public boolean isBalanced(TreeNode root) {
return height(root) >= 0;
}

private int height(TreeNode root) {
if (root == null) {
return 0;
}
int l = height(root.left);
int r = height(root.right);
if (l == -1 || r == -1 || Math.abs(l - r) > 1) {
return -1;
}
return 1 + Math.max(l, r);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode* root) {
function<int(TreeNode*)> height = [&](TreeNode* root) {
if (!root) {
return 0;
}
int l = height(root->left);
int r = height(root->right);
if (l == -1 || r == -1 || abs(l - r) > 1) {
return -1;
}
return 1 + max(l, r);
};
return height(root) >= 0;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def isBalanced(self, root: Optional[TreeNode]) -> bool:
def height(root):
if root is None:
return 0
l, r = height(root.left), height(root.right)
if l == -1 or r == -1 or abs(l - r) > 1:
return -1
return 1 + max(l, r)

return height(root) >= 0

• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func isBalanced(root *TreeNode) bool {
var height func(*TreeNode) int
height = func(root *TreeNode) int {
if root == nil {
return 0
}
l, r := height(root.Left), height(root.Right)
if l == -1 || r == -1 || abs(l-r) > 1 {
return -1
}
if l > r {
return 1 + l
}
return 1 + r
}
return height(root) >= 0
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function isBalanced(root: TreeNode | null): boolean {
const dfs = (root: TreeNode | null) => {
if (root == null) {
return 0;
}
const left = dfs(root.left);
const right = dfs(root.right);
if (left === -1 || right === -1 || Math.abs(left - right) > 1) {
return -1;
}
return 1 + Math.max(left, right);
};
return dfs(root) > -1;
}

• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isBalanced = function (root) {
const height = root => {
if (!root) {
return 0;
}
const l = height(root.left);
const r = height(root.right);
if (l == -1 || r == -1 || Math.abs(l - r) > 1) {
return -1;
}
return 1 + Math.max(l, r);
};
return height(root) >= 0;
};

• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn is_balanced(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
Self::dfs(&root) > -1
}

fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let left = Self::dfs(&node.left);
let right = Self::dfs(&node.right);
if left == -1 || right == -1 || (left - right).abs() > 1 {
return -1;
}
1 + left.max(right)
}
}