Balanced Binary Tree

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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / 
  9  20
    /  
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / 
     2   2
    / 
   3   3
  / 
 4   4

Return false.

Approach #1: C++.[recursive]

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if (root == nullptr) return true;
        
        int left = helper(root->left);
        int right = helper(root->right);
        
        return abs(left - right) <= 1 && isBalanced(root->left) && isBalanced(root->right);
    }
    
private:
    int helper(TreeNode* root) {
        if (root == nullptr) return 0;
        return max(helper(root->left), helper(root->right)) + 1;
    }
};

　　

Approach #2: Java.[DFS]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        return dfs(root) != -1;
    }
    
    private int dfs(TreeNode root) {
        if (root == null) return 0;
        
        int left = dfs(root.left);
        if (left == -1) return -1;
        int right = dfs(root.right);
        if (right == -1) return -1;
        
        if (Math.abs(left - right) > 1) return -1;
        
        return Math.max(left, right) + 1;
    }
        
}

　　

Approach #3: Python.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        return abs(self.solve(root.left) - self.solve(root.right)) and self.isBalanced(root.left) and self.isBalanced(root.right)
    
    def solve(self, root):
        if not root:
            return 0
        return max(self.solve(root.left), self.solve(root.right)) + 1