二叉查找树的个数，BT search 1, 二叉树遍历的非递归形式，Flatten Binary Tree to Linked List

主要是卡特兰特数的递归公式：

n>1

h(0) = 1;

0<=i<=n-1;左边是i个节点，右边是n-i-1个节点

h(n) += h(i) * h(n-i-1);

class Solution {public:    int numTrees(int n) {        if(n == 0)            return 0;                int* num = new int[n+1];        num[0] = 1;        num[1] = 1;        for(int i = 2; i <= n; ++ i)        {            num[i] = 0;            for(int j = 0; j <= i-1; ++ j)                num[i] += num[j] * num[i-j-1];        }        return num[n];    }};

二叉树的先序遍历的非递归形式

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {            stack<TreeNode*> mystack;    vector<int> num;    if(root == NULL) return num;    mystack.push(root);    TreeNode* p;    while(!mystack.empty())    {    p = mystack.top();    mystack.pop();    num.push_back(p->val);    if(p->right) mystack.push(p->right);    if(p->left) mystack.push(p->left);    }    return num;    }};

中序遍历

出现的错误应该是因为，栈中没有元素却出栈，所以不对，应该是两个判断条件root != NULL || !mystack.empty()

class Solution {public:    vector<int> inorderTraversal(TreeNode *root) {        stack<TreeNode*> mystack;vector<int> num;if(root == NULL) return num;while(root != NULL || !mystack.empty()){for(; root ; root = root->left)mystack.push(root);root = mystack.top();num.push_back(root->val);mystack.pop();root = root->right;}return num;    }};

后续遍历

需要注意的是，左右根的遍历，对于输出的节点需要满足两点 1）是叶子节点  2）其孩子都已经输出， 还有就是取节点，但节点暂时不出栈，等到条件满足再出栈

class Solution {public:    vector<int> postorderTraversal(TreeNode *root) {        stack<TreeNode*> mystack;        vector<int> num;        if(root == NULL) return num;        mystack.push(root);        TreeNode* p=root;        TreeNode* out = root;        while(!mystack.empty())        {        p = mystack.top();        if((p->left == NULL && p->right == NULL) || p->left == out || p->right == out)        {        mystack.pop();        num.push_back(p->val);        out = p;        }else        {        if(p->right) mystack.push(p->right);        if(p->left) mystack.push(p->left);        }        }                        return num;    }};

二叉树的层次遍历，只不过每层放入一个vector中，就是做到层次区分

vector<vector<int> > levelOrder(TreeNode* root){vector<vector<int> > result;vector<int> temp;queue<TreeNode* > myQueue;TreeNode* p=root;if(root == NULL) return result;myQueue.push(root);int pre = 1;int cur = 0;while(!myQueue.empty()){p = myQueue.front();temp.push_back(p->val);myQueue.pop();--pre;if(p->left != NULL){myQueue.push(p->left);++ cur;}if(p->right != NULL){myQueue.push(p->right);++ cur;}if(pre == 0){result.push_back(temp);temp.clear();pre = cur;cur = 0;}}return result;}

Flatten Binary Tree to Linked List

主要就是先序遍历，不过不能是递归的，因为递归只是返回根节点，我需要的是最后一个节点和跟节点，所以用迭代的先序遍历方式 

class Solution {public:    TreeNode* flattenTree(TreeNode* node){if(node == NULL)return NULL;if(node->left == NULL && node->right == NULL)return node;stack<TreeNode* > nodeStack;TreeNode *temp = NULL, *head = node;if(node->right!=NULL)nodeStack.push(node->right);if(node->left!=NULL)nodeStack.push(node->left);while(!nodeStack.empty()){temp = nodeStack.top();nodeStack.pop();node->right = temp;node->left = NULL;node = node->right;if(temp->right!=NULL)nodeStack.push(temp->right);if(temp->left!=NULL)nodeStack.push(temp->left);}return head;}void flatten(TreeNode *root) {if(root == NULL ||root->left == NULL && root->right == NULL)return;//树的先序遍历转化为链接在节点的右子树上的链表flattenTree(root);}};