110. Balanced Binary Tree

Easy

Problem Description

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

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

Examples

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
Jump to Solution: Python Java C++ JavaScript C#

Python Solution


# 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
def isBalanced(root: Optional[TreeNode]) -> bool:
    def checkHeight(node: Optional[TreeNode]) -> int:
        if not node:
            return 0
        
        # Get heights of left and right subtrees
        leftHeight = checkHeight(node.left)
        if leftHeight == -1:
            return -1
        
        rightHeight = checkHeight(node.right)
        if rightHeight == -1:
            return -1
        
        # Check if current node is balanced
        if abs(leftHeight - rightHeight) > 1:
            return -1
        
        # Return height of current subtree
        return max(leftHeight, rightHeight) + 1
    
    return checkHeight(root) != -1

Java Solution


/**
 * 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 {
    private int checkHeight(TreeNode node) {
        if (node == null) {
            return 0;
        }
        
        // Get heights of left and right subtrees
        int leftHeight = checkHeight(node.left);
        if (leftHeight == -1) {
            return -1;
        }
        
        int rightHeight = checkHeight(node.right);
        if (rightHeight == -1) {
            return -1;
        }
        
        // Check if current node is balanced
        if (Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        }
        
        // Return height of current subtree
        return Math.max(leftHeight, rightHeight) + 1;
    }
    
    public boolean isBalanced(TreeNode root) {
        return checkHeight(root) != -1;
    }
}

C++ Solution


/**
 * 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 {
private:
    int checkHeight(TreeNode* node) {
        if (!node) {
            return 0;
        }
        
        // Get heights of left and right subtrees
        int leftHeight = checkHeight(node->left);
        if (leftHeight == -1) {
            return -1;
        }
        
        int rightHeight = checkHeight(node->right);
        if (rightHeight == -1) {
            return -1;
        }
        
        // Check if current node is balanced
        if (abs(leftHeight - rightHeight) > 1) {
            return -1;
        }
        
        // Return height of current subtree
        return max(leftHeight, rightHeight) + 1;
    }
    
public:
    bool isBalanced(TreeNode* root) {
        return checkHeight(root) != -1;
    }
};

JavaScript Solution


/**
 * 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 checkHeight = (node) => {
        if (!node) {
            return 0;
        }
        
        // Get heights of left and right subtrees
        const leftHeight = checkHeight(node.left);
        if (leftHeight === -1) {
            return -1;
        }
        
        const rightHeight = checkHeight(node.right);
        if (rightHeight === -1) {
            return -1;
        }
        
        // Check if current node is balanced
        if (Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        }
        
        // Return height of current subtree
        return Math.max(leftHeight, rightHeight) + 1;
    };
    
    return checkHeight(root) !== -1;
};

C# Solution


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private int CheckHeight(TreeNode node) {
        if (node == null) {
            return 0;
        }
        
        // Get heights of left and right subtrees
        int leftHeight = CheckHeight(node.left);
        if (leftHeight == -1) {
            return -1;
        }
        
        int rightHeight = CheckHeight(node.right);
        if (rightHeight == -1) {
            return -1;
        }
        
        // Check if current node is balanced
        if (Math.Abs(leftHeight - rightHeight) > 1) {
            return -1;
        }
        
        // Return height of current subtree
        return Math.Max(leftHeight, rightHeight) + 1;
    }
    
    public bool IsBalanced(TreeNode root) {
        return CheckHeight(root) != -1;
    }
}

Complexity Analysis

Solution Explanation

This solution uses a bottom-up approach:

Key points: