#129求根到叶数字之和

回溯放进vector，然后从后往前拿，乘1 10 100 ... 很基础的回溯

my code：

    void backtrack(int depth, TreeNode* cur, vector<TreeNode*> &vec, int &sum){if(cur->left==nullptr &&cur->right==nullptr){size_t a=1;for(int i=vec.size()-1;i>=0;i--){sum+=vec[i]->val*a;a*=10;}return;}if(cur->left){vec.push_back(cur->left);backtrack(depth+1,cur->left, vec, sum);vec.pop_back();}if(cur->right){vec.push_back(cur->right);backtrack(depth+1,cur->right,vec, sum);vec.pop_back();}}int sumNumbers(TreeNode* root) {int sum=0;vector<TreeNode*> vec={root};backtrack(1,root,vec,sum);return sum;}

随想录跟我思路差不多，但这两个是放globa的：int result; vector<int> path; 最近总觉得global不安全不想放 

#1382二叉树变平衡

本来遇到这种会换root的就头疼，然后看了眼随想录，是要先变成vec，再从vec重构，就会了

mycode：

    vector<int> vec;void inorder(TreeNode* node){if(node==nullptr) return;inorder(node->left);vec.push_back(node->val);inorder(node->right);}TreeNode* build_tree(int start, int end){if(start>end) return nullptr;TreeNode* root= new TreeNode();int mid=(start+end)/2;int val=vec[mid];root->val=val;root->left=build_tree(start, mid-1);root->right=build_tree(mid+1, end);return root;}TreeNode* balanceBST(TreeNode* root) {inorder(root);return build_tree(0,vec.size()-1);}

#100相同的树 easy

以任何方式遍历比较即可：和随想录一模一样

    bool traverse(TreeNode* n1, TreeNode* n2){if(n1==nullptr && n2==nullptr) return true;if(n1==nullptr && n2) return false;if(n1 && n2==nullptr) return false;if(n1->val!=n2->val) return false;return traverse(n1->left, n2->left) && traverse(n1->right, n2->right);}bool isSameTree(TreeNode* p, TreeNode* q) {return traverse(p,q);}