代码随想录算法Day17 | 110.平衡二叉树 、257. 二叉树的所有路径 、404.左叶子之和

110.平衡二叉树

复习一下递归三部曲：

1.确定递归函数的参数和返回值

2.确认终止条件

3.确认单层递归的逻辑

此题为比较高度，使用左右中后序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    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 result;
        if(abs(rightheight-leftheight)>1)result=-1;
        else{
            result=1+max(leftheight,rightheight);
        }
        return result;
    }
    bool isBalanced(TreeNode* root) {
        return getheight(root)==-1?false:true;
    }
};

257. 二叉树的所有路径

回溯和递归是相辅相成的，只要有递归就一定有回溯

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void traversal(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){
            traversal(cur->left,path,result);
            path.pop_back();
        }
        if(cur->right){
            traversal(cur->right,path,result);
            path.pop_back(); 
        }
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string>result;
        vector<int> path;
        if(root==NULL)return result;
        traversal(root,path,result);
        return result;
    }
};

404.左叶子之和

1.确定递归函数的参数和返回值

判断一个树的左叶子节点之和，那么一定要传入树的根节点，递归函数的返回值为数值之和，所以为int

2.确定终止条件

如果遍历到空节点，那么左叶子值一定是0

3.确定单层递归的逻辑

当遇到左叶子节点的时候，记录数值，然后通过递归求取左子树左叶子之和，和右子树左叶子之和，相加便是整个树的左叶子之和。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if(root == NULL)return 0;
        if(root->left==NULL&&root->right==NULL)return 0;
        int leftValue = sumOfLeftLeaves(root->left);
        if(root->left&&!root->left->left&&!root->left->right){
            leftValue = root->left->val;
        }
        int rightValue = sumOfLeftLeaves(root->right);
        int sum = leftValue+rightValue;
        return sum;
    }
};