110:Balanced Binary Tree【樹】【DFS】

題目鏈接：click~

/*題意：判斷一顆平衡二叉樹是否平衡*/

/**
 *思路：平衡二叉樹的每個結點的左右子樹的深度差不超過1
 *      DFS記錄每個結點的左右子樹的深度，判斷是否平衡
 */

class Solution {

public:
    bool Judge(TreeNode *T, int &dep) {
        if(T == NULL) {
            dep = 0;
            return true;
        }

        int leftdepth, rightdepth;
        bool leftBalance = Judge(T->left, leftdepth);
        bool rightBalance = Judge(T->right, rightdepth);

        dep = max(leftdepth, rightdepth) + 1;

        return leftBalance && rightBalance && (abs(leftdepth-rightdepth) <= 1);
    }
    bool isBalanced(TreeNode *root) {
        int dep;
        return Judge(root, dep);
    }
};