【樹】高度平衡二叉樹的判定

題目：

 

 

解答：

平衡二叉樹要求左子樹和右子樹的高度相差為1，且左右子樹都是平衡二叉樹，顯然要計算二叉樹的高度的函數。

/**
 * 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:
    // 求深度
    int depthTree(TreeNode *root)
    {
        int depth = 0;
        if (root)
        {
            int leftdepth = depthTree(root->left);
            int rightdepth = depthTree(root->right);

            depth = leftdepth > rightdepth ? (leftdepth + 1) : (rightdepth + 1);
        }
        return depth;
    }

    bool isBalanced(TreeNode* root) 
    {
        if (NULL == root)
        {
            return true;
        }

        int leftdepth = depthTree(root->left);
        int rightdepth = depthTree(root->right);

        if (abs(leftdepth - rightdepth) > 1)
        {
            return false;
        }
        else
        {
            return isBalanced(root->left) && isBalanced(root->right);
        }
    }
};