110. Balanced Binary Tree

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

根據題意，左側樹高和右側樹高的差如果大於 1 即沒有平衡，使用遞迴解。

/**
 * 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) {
        if(!root) return true;
        
        // 走訪左邊的樹
        int left = getHeight(root->left);
        
        // 走訪右邊的樹
        int right = getHeight(root->right);

        return abs(right - left) < 2 && isBalanced(root->left) && isBalanced(root->right);
    }
    int getHeight(TreeNode* root) {
        if(!root) return -1;
        // 每遞迴一次，height + 1
        return 1 + max(getHeight(root->left), getHeight(root->right));
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(N)$