剑指offer系列——39.平衡二叉树

Q：输入一棵二叉树，判断该二叉树是否是平衡二叉树。
A：
结合上一题的计算树的高度。

    bool IsBalanced_Solution(TreeNode *pRoot) {
        vector<int> diff = {-1, 0, 1};
        if (pRoot == nullptr)
            return true;
        int l = TreeDepth(pRoot->left);
        int r = TreeDepth(pRoot->right);
        int d = l - r;
        vector<int>::iterator it;
        it = find(diff.begin(), diff.end(), d);
        if (it == diff.end())
            return false;
        else {
            bool left = IsBalanced_Solution(pRoot->left);
            bool right = IsBalanced_Solution(pRoot->right);
            return left && right;
        }
    }

    int TreeDepth(TreeNode *pRoot) {
        if (pRoot == nullptr)
            return 0;
        int l = TreeDepth(pRoot->left);
        int r = TreeDepth(pRoot->right);
        return l > r ? l + 1 : r + 1;
    }

但这样做有一个问题，就是下层一直累积遍历很多次。这样直接在遍历过程中进行判断。

    bool IsBalanced(TreeNode *root, int & dep){
        if(root == NULL){
            return true;
        }
        int left = 0;
        int right = 0;
        if(IsBalanced(root->left,left) && IsBalanced(root->right, right)){
            int dif = left - right;
            if(dif<-1 || dif >1)
                return false;
            dep = (left > right ? left : right) + 1;
            return true;
        }
        return false;
    }
    bool IsBalanced_Solution(TreeNode* pRoot) {
        int dep = 0;
        return IsBalanced(pRoot, dep);
    }