Day 16 迎财神

坚持如此hard

玄之又玄，众妙之门

一、理论知识

回顾【深度】和【高度】的概念，现在主要还是写递归

2）

初识【回溯】

3）左叶子的明确定义：节点A的左孩子不为空，且左孩子的左右孩子都为空（说明是叶子节点），那么A节点的左孩子为左叶子节点

二、刷题学习

110. 平衡二叉树

实则在求高度。

/*** 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* root){//1if(root==nullptr){return 0;}//2int left=getheight(root->left);if(left==-1) return -1;int right=getheight(root->right);if(right==-1) return -1;//3int result=abs(left-right);if(result>1) return -1;else return max(left,right)+1;}bool isBalanced(TreeNode* root) {int ans=getheight(root);if(ans==-1) return false;else return true;}
};

257. 二叉树的所有路径

/*** 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 traverse(TreeNode* root,vector<int>&path,vector<string>&result){//midpath.push_back(root->val);//stopif(root->left==nullptr && root->right==nullptr) {//收割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;}//leftif(root->left){traverse(root->left,path,result);path.pop_back();}if(root->right){traverse(root->right,path,result);path.pop_back();}}vector<string> binaryTreePaths(TreeNode* root) {vector<string> result;vector<int> path;if (root == NULL) return result;traverse(root, path, result);return result;}
};

404. 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) {//1if(root==nullptr) return 0;if (root->left == NULL && root->right== NULL) return 0;int left= sumOfLeftLeaves(root->left);if (root->left && !root->left->left && !root->left->right) { // 左子树就是一个左叶子的情况left = root->left->val;}int right=sumOfLeftLeaves(root->right); return left+right;}
};

三、新年快乐

道可道，非常道