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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)
    }
}

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