[LeetCode]Balanced Binary Tree-优快云博客

本文介绍了一种用于判断二叉树是否为平衡二叉树的算法实现。通过递归方式计算二叉树各子树的高度差，若任意两棵子树的高度差不超过1，则认为该二叉树为平衡二叉树。文章提供了完整的C++代码示例。

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
	//check if it is a balanced tree
	//so we check the diff between the max height of each branch
	//all compare can be done along the way down to calculate max height 
public:
	int CheckHeight(TreeNode* root, bool& ans)
	{
		if(!root || !ans)//if is is not a balance tree already, we just return
			return 0;
		int leftHeight = CheckHeight(root->left, ans);
		int rightHeight = CheckHeight(root->right, ans);
		if(abs(leftHeight-rightHeight) > 1)
			ans = false;
		return max(leftHeight, rightHeight)+1;
	}
    bool isBalanced(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        //get the height of each subtree
		bool ans = true;
		CheckHeight(root, ans);
		return ans;
    }
};

second time

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int isBalancedUtil(TreeNode* root, bool& isValid)
    {
        if(root == NULL) return 0;
        if(!isValid) return -1;
        int leftHeight = isBalancedUtil(root->left, isValid);
        int rightHeight = isBalancedUtil(root->right, isValid);
        if(abs(leftHeight-rightHeight) > 1) isValid = false;
        return max(leftHeight, rightHeight)+1;
    }
    bool isBalanced(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        bool ans = true;
        isBalancedUtil(root, ans);
        return ans;
    }
};