代码随想录算法训练营第十七天|LeetCode110 平衡二叉树、LeetCode257 二叉树的所有路径

R.S.G.久夏2024-05-25 5:04

题1：

指路：LeetCode110 平衡二叉树

思路与代码：

左右子树的高度差小于等于1。对于这个题，递归比迭代方便太多，我也想过迭代，但是我没有写出来，大家可以自己试一下。递归代码如下：

cpp 复制代码

class Solution {
public:
    //递归
    int getHeight  (TreeNode* node) {
        if (node == NULL) {return 0;}
        int leftHeight = getHeight(node->left);
        if (leftHeight == -1) return -1;
        int rightHeight = getHeight(node->right);
        if (rightHeight == -1) return -1;
        int ans = abs(leftHeight - rightHeight);
        if (ans > 1) return -1;  // 绝对值超过1符合条件
        else return 1 + max(leftHeight, rightHeight);
       /* return abs(leftHeight - rightHeight) > 1 ? -1 : 1 + max(leftHeight, rightHeight);*/
    } 
    bool isBalanced(TreeNode* root) {
    if (getHeight(root)== -1) 
    return false;
    return true;
    /*return getHeight(root) == -1 ? false : true;*/
    }
};

题2：

指路：LeetCode257 二叉树的所有路径

思路与代码：

递归进行前序遍历，找到子节点记录路径之后回溯回退路径。我还没会呢，先看看代码吧。

cpp 复制代码

class Solution {
private:
    void treversal(TreeNode* cur, vector<int>& path, vector<string>& result) {
        path.push_back(cur->val);
        if (cur->left == NULL && cur->right == NULL) {
            string sPath;
            for (int i = 0; i < path.size() - 1; i++) {
                sPath += to_string(path[i]);
                sPath += "->";
            }
            sPath += to_string(path[path.size() - 1]);
            result.push_back(sPath);
            return ;
        }
        if (cur->left) {
            treversal(cur->left, path, result);
            path.pop_back();
        }
        if (cur->right) {
            treversal(cur->right, path, result);
            path.pop_back();
        }
    }
    public:
    vector<string> binaryTreePaths(TreeNode* root) {
    vector<string> result;
    vector<int> path;
    if (root == NULL) return result;
    treversal(root, path, result);
    return result;
    }
};