[LeetCode] 105 & 106

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105. Construct Binary Tree from Preorder and Inorder Traversal

题目：给定一个树的preOrder 和 inOrder 表达，返回这个树

思路：preOrder的第一个数字就是root，然后去inOrder的数组里找root的位置，这个位置的左边为左子树，右边为右子树，然后递归即可

class Solution {public:    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        return helper(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);    }        TreeNode* helper(vector<int>& preorder, int pS, int pE, vector<int>& inorder, int iS, int iE)    {        if(preorder.size() != inorder.size())         {            cout<<":dfjaksdfjak"<<endl;            return NULL;        }                if(pS > pE || iS > iE) return NULL;                if(pS == pE || iS == iE)        {            return new TreeNode(preorder[pS]);        }                int rootVal = preorder[pS];        //cout<<"root->va: "<<rootVal;        TreeNode* root = new TreeNode(rootVal);        int i = 0;        for(i = iS; i <= iE; i++)        {            if(inorder[i] == rootVal) break;        }                int length = i - iS;        //cout<<" length of left subtree is: "<<length<<endl;                root->left = helper(preorder, pS+1, pS+length, inorder, iS, i-1);        root->right = helper(preorder, pS+length+1, pE, inorder, i+1, iE);                return root;    }};

106. Construct Binary Tree from Inorder and Postorder Traversal

题目：给定一个树的inOrder 和 postOrder 表达，返回这个树

思路：和105基本相同，postOrder的最后一个数为root，然后去inOrder的数组里找root的位置，这个位置的左边为左子树，右边为右子树，然后递归即可

class Solution {public:    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {        return helper(postorder, 0, postorder.size()-1, inorder, 0, inorder.size()-1);    }        TreeNode* helper(vector<int>& postorder, int pS, int pE, vector<int>& inorder, int iS, int iE)    {        if(postorder.size() != inorder.size())         {            cout<<":dfjaksdfjak"<<endl;            return NULL;        }                if(pS > pE || iS > iE) return NULL;                if(pS == pE || iS == iE)        {            return new TreeNode(postorder[pS]);        }                int rootVal = postorder[pE];        //cout<<"root->va: "<<rootVal;        TreeNode* root = new TreeNode(rootVal);        int i = 0;        for(i = iS; i <= iE; i++)        {            if(inorder[i] == rootVal) break;        }                int length = i - iS;        //cout<<" length of left subtree is: "<<length<<endl;                root->left = helper(postorder, pS, pS+length-1, inorder, iS, i-1);        root->right = helper(postorder, pS+length, pE-1, inorder, i+1, iE);                return root;    }};

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