LeetCode #110: Balanced Binary Tree (C/C++).

easy

source: https://leetcode.com/problems/balanced-binary-tree
C/C++ Solution to LeetCode problem 110. Balanced Binary Tree.

Problem

---

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

Constraints

---

• The number of nodes in the tree is in the range [0, 5000].
• -104 <= Node.val <= 104

Solution

---

We will use recursion to perform a DFS (Depth First Search) in order to determine the heigh of each subtree.

• We call the recursive method with the root node.
  • If the current node doesn’t exist, return 0.
  • Explore the left and right nodes.
  • If the absolute difference of nodes’ depth is higher than 1, we return -1.
  • If the the depth of the left or right node is -1, we return -1 for the current node.
  • Otherwise, we return the max hight of the nodes (left vs right) plus 1.
• If the final result == -1 then the tree is unbalanced.

/**
 * 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 exploreNode(TreeNode* node) {
    if (node == nullptr)  return 0;

    int lH = exploreNode(node->left);
    int rH = exploreNode(node->right);
  
    if (lH == -1 || rH == -1 || abs(lH - rH) > 1) 
      return -1;

    return max(lH, rH) + 1;
 }

public:
  bool isBalanced(TreeNode* root) {
    int r =  exploreNode(root); 
    return  r != -1;
  }
};