LeetCode-Balanced Binary Tree

平衡二叉树的判断，DFS，

一直没想到怎么同时返回结点的长度和状态，把高度放到参数里面比较简便！

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int height = 0;
        return getHeight(root, height);
        
    }
    bool getHeight(TreeNode *root, int &height) {
        if (!root) {
            return true;
        }
        int leftheight = 0;
        int rightheight = 0;
        if (getHeight(root->left, leftheight) && getHeight(root->right, rightheight)) {
            height = max(leftheight, rightheight) + 1;
            if (abs(leftheight - rightheight) <= 1) {
                return true;
            }
        }
        return false;
    }
};

 不过其实不需要用两个变量来表示状态，可以直接用一个int返回值来表示，

 如果是false的则直接返回-1，否则返回高度：

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int ret = getHeight(root);
        if (ret == -1) {
            return false;
        }
        return true;
    }
    int getHeight(TreeNode *root) {
        if (root == NULL) {
            return 0;
        }
        int left = getHeight(root->left);
        int right = getHeight(root->right);
        if (left == -1 || right == - 1) {
            return -1;
        }
        if (abs(left - right) > 1) {
            return -1;
        }
        return (max(left, right) + 1);
    }
};