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.

Most of the Binary tree problems can be tackled by Recursion – because the tree itself is defined recursively. Depending on the directions, we can perform recursions either top-down (from the root to the leaves) or the bottom-up (from the leaves to the root). In this particular problem, the bottom-up approach is more efficient as the depths of the trees are passed up without duplicate computation efforts.

Top-Down Recursion Algorithm to Validate a Balanced Binary Tree by Checking the Depths

We first define a recursive function to get the depth of a binary tree. Then, at each node from root to the leaves, we check if the depth of its left branch and the right subtree is no more than 1 difference. Then, we also need to recursively check if the sub-trees are balanced.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

/**
 * 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) return true;
        int left = maxDepth(root->left);
        int right = maxDepth(root->right);
        if (abs(left - right) > 1) return false;
        return (isBalanced(root->left) && isBalanced(root->right));            
    }
 
private:
    int maxDepth(TreeNode* root) {        
        if (!root) return true;
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
    }
};

/**
 * 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) return true;
        int left = maxDepth(root->left);
        int right = maxDepth(root->right);
        if (abs(left - right) > 1) return false;
        return (isBalanced(root->left) && isBalanced(root->right));            
    }

private:
    int maxDepth(TreeNode* root) {        
        if (!root) return true;
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
    }
};

The time complexity is O(N^2) as at each node, the depths are recalculated repeatedly. It is worse when the binary tree is degenerated into a linked-list.

Bottom-up Recursion Algorithm to Validate a Balanced Binary Tree by Passing Up the Depths

We can compute the depth for the binary sub-tree, and pass it up. If the tree is un-balanced, we pass the value as -1, then we don’t need to re-calculate the depths for a upper-level nodes (parent nodes) because the entire tree will be un-balanced anyway.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

/**
 * 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) {
        return depth(root) != -1;
    }
private:
    int depth(TreeNode* root) {
        if (root == nullptr) return 0;
        int left = depth(root->left);
        if (left == -1) return -1;
        int right = depth(root->right);
        if (right == -1) return -1;
        if (abs(left - right) > 1) return -1;
        return max(right, left) + 1;
    }
};

/**
 * 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) {
        return depth(root) != -1;
    }
private:
    int depth(TreeNode* root) {
        if (root == nullptr) return 0;
        int left = depth(root->left);
        if (left == -1) return -1;
        int right = depth(root->right);
        if (right == -1) return -1;
        if (abs(left - right) > 1) return -1;
        return max(right, left) + 1;
    }
};

The time complexity is improved from O(N^2) to O(N). Both recursion approaches require O(N) stack space.

–EOF (The Ultimate Computing & Technology Blog) —