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最新推荐文章于 2018-04-04 17:51:00 发布

原创最新推荐文章于 2018-04-04 17:51:00 发布 · 423 阅读

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本文介绍了一种检查二叉树是否平衡的方法。通过递归计算每个节点的左右子树高度差，若所有节点的高度差不超过1，则认为该二叉树是平衡的。文章提供了一个具体的C++实现示例。

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4.1 Implement a function to check if a binary tree is balanced. For the purpose of this question, a balanced tree is defined to be a tree such that the height of the two subtrees of any node never differ by more than one.

struct TreeNode {
    int data;
    TreeNode *left, *right;
    TreeNode(int v) : data(v), left(NULL), right(NULL) {}
};

int checkHeight(TreeNode *node) {
    if (node == NULL) {
        return 0;
    }
    
    int leftHeight = checkHeight(node->left);
    if (leftHeight == -1) {
        return -1;
    }
    
    int rightHeight = checkHeight(node->right);
    if (rightHeight == -1) {
        return -1;
    }
    
    if (abs(leftHeight - rightHeight) > 1) {
        return -1;
    } else {
        return max(leftHeight, rightHeight) + 1;
    }
}

bool isBalanced(TreeNode *node) {
    return checkHeight(node) != -1;
}