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.
PythonJavaC++GoTypeScriptRustJavaScript
# 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(code-box)
/**
* 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);
}
}(code-box)
/**
* 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;
}
};(code-box)
/**
* 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
}(code-box)
/**
* 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;
}(code-box)
// 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::cell::RefCell;
use std::rc::Rc;
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)
}
}(code-box)
/**
* 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;
};(code-box)
