[Leetcode] 105, 106, 96

105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

Solution: 直接递归求解。

关于先序遍历、中序遍历详见：http://blog.csdn.net/prince_jun/article/details/7699024

Code:

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        return buildTree(preorder.begin(), inorder.begin(), inorder.size());    }private:    TreeNode* buildTree(vector<int>::iterator prebegin, vector<int>::iterator inbegin, int size) {        if(size<=0) return NULL;        TreeNode* root = new TreeNode(*prebegin);        vector<int>::iterator itroot = find(inbegin, inbegin+size, *prebegin);        int sizeleft = itroot-inbegin;        int sizeright = size - 1 - sizeleft;         TreeNode* left = buildTree(prebegin+1, inbegin, sizeleft);        TreeNode* right = buildTree(prebegin+sizeleft+1, itroot+1, sizeright);        root->left = left;        root->right = right;        return root;    }};

106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

Solution: 递归。

Code:

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {        return buildTree(inorder.begin(), postorder.end()-1, postorder.size());    }private:    TreeNode* buildTree(vector<int>::iterator inbegin, vector<int>::iterator postend, int size){        if(size==0) return NULL;        vector<int>::iterator itroot = find(inbegin, inbegin+size, *postend);        TreeNode* root = new TreeNode(*postend);                int sizeleft = itroot - inbegin;        int sizeright = size - 1 - sizeleft;        TreeNode* left = buildTree(inbegin, postend-1-sizeright, sizeleft);        TreeNode* right = buildTree(itroot+1, postend-1, sizeright);                root->left = left;        root->right = right;        return root;    }};

96. Unique Binary Search Trees

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.

   1         3     3      2      1    \       /     /      / \      \     3     2     1      1   3      2    /     /       \                 \   2     1         2                 3

Solution:输入n时，可以分为根节点是1-n的n种子情况。对于根节点是x时，左子树是比x小的点，一共有x-1个不相等的数，其可能性的数量等于numTrees(x-1)；右子树是比x大的点，一共有n-x个点，其可能性等于numTrees(n-x)。

根据上面的分析可以得到递推公式：

根据这条递推公式就能解除答案，但要注意，递归会超时，这题只能用迭代来写。

Code:

class Solution {public:    int numTrees(int n) {        vector<int> ans(n+1,0);        ans[0] = 1;        ans[1] = 1;        int t = 1;        while(t<n){            t++;            for(int i=1; i<=t; i++){                ans[t] += ans[i-1]*ans[t-i];            }        }        return ans[n];    }};