110. 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.

思路

思路很简单，递归判断左右子数的深度差是否在1以内即可。

方法一：

• 构建一个height函数，递归计算树的深度；
• 在主函数中判断节点的左右子树深度是否大于1，成立则返回false，不成立继续递归判断该节点的左右子树。

方法二：

方法一种用了两次递归判断，这里进一步优化：只要左右子树高度差大于1就直接返回false，不继续递归计算了。

c++11

方法一：

/**
 * 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;
        if (std::abs(height(root->left) - height(root->right)) > 1) return false;
        return isBalanced(root->left) and isBalanced(root->right);
    }

    int height(TreeNode* root) {
        if (!root) return 0;
        return std::max(height(root->left), height(root->right)) + 1;
    }
};

方法二：

/**
 * 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;
        if (height(root) == -1) return false;
        return true;
    }

    int height(TreeNode* root) {
        if (!root) return 0;

        int left_height = height(root->left);
        if (left_height == -1) return -1;
        int right_height = height(root->right);
        if (right_height == -1) return -1;

        if (std::abs(left_height - right_height) > 1) return -1;
        return std::max(left_height, right_height) + 1;
    }
};