【Construct Binary Tree from Preorder and Inorder Traversal】cpp

题目：

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

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

代码：

/**
 * 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) {
        if ( preorder.size()== || inorder.size()== ) return NULL;
        return Solution::buildTreePI(preorder, , preorder.size()-, inorder, , inorder.size()-);
    }
        static TreeNode* buildTreePI(
            vector<int>& preorder,
            int beginP, int endP,
            vector<int>& inorder,
            int beginI, int endI)
        {
            // terminal condition & corner case
            if ( beginP>endP ) return NULL;
            // resurisve process
            TreeNode *root = new TreeNode(-);
            root->val = preorder[beginP];
            // find the root node in inorder traversal
            int rootPosInorder = beginI;
            for ( int i = beginI; i <= endI; ++i )
            {
                if ( inorder[i]==root->val ) { rootPosInorder=i; break;}
            }
            int leftSize = rootPosInorder - beginI;
            int rightSize = endI - rootPosInorder;
            root->left = Solution::buildTreePI(preorder, beginP+, beginP+leftSize, inorder, beginI, rootPosInorder-);
            root->right = Solution::buildTreePI(preorder, endP-rightSize+, endP, inorder, rootPosInorder+, endI);
            return root;
        }
};

tips：

经典题目。直接学习高手代码

http://fisherlei.blogspot.sg/2013/01/leetcode-construct-binary-tree-from.html

http://bangbingsyb.blogspot.sg/2014/11/leetcode-construct-binary-tree-from.html

这里注意在递归传递vector元素下标的时候，一定是绝对下标（一开始疏忽写成了相对下标，debug了不少时间）

=========================================

第二次过这道题，大体思路还记得，顺着思路摸了下来，代码改了一次AC了。注意每次要找到的是preorder[bp]而不是preorder[0]。

/**
 * 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 Solution::build(preorder, , preorder.size()-, inorder, , inorder.size()-);
        }
        TreeNode* build(
            vector<int>& preorder, int bp, int ep,
            vector<int>& inorder, int bi, int ei
            )
        {
            if ( bp>ep || bi>ei ) return NULL;
            TreeNode* root = new TreeNode(preorder[bp]);
            int pos = Solution::findPos(inorder, bi, ei, preorder[bp]);
            int left_range = pos - bi;
            root->left = Solution::build(preorder, bp+, bp+left_range, inorder, bi, bi+left_range-);
            root->right = Solution::build(preorder, bp+left_range+, ep, inorder, bi+left_range+, ei);
            return root;
        }
        int findPos(vector<int>& order, int begin, int end, int val)
        {
            for ( int i=begin; i<=end; ++i ) if (order[i]==val) return i;
        }
};