构建二叉树（前中序构建105，中后序构建106）--非递归的方法值得研究

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一、前中序构建
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.
题目大意：根据二叉树的前序遍历和中序遍历重新构建一颗二叉树
方法一、递归的方法构建
思路：由于前序遍历的第一个数字为根节点的数字，所以构建根节点，然后去中序遍历中找到根节点的位置，从而定位出根节点的左子树所拥有的孩子节点的数目，以及根节点的右子树的节点数，从而递归的构建左右子树

 TreeNode* buildTree(vector<int>& preorder,int begin1,int end1,vector<int>& inorder,int begin2,int end2)    {        if(begin1 > end1)            return NULL;        else if (begin1 == end1)            return new TreeNode(preorder[begin1]);        TreeNode* root = new TreeNode(preorder[begin1]);        int i = begin2;        for(; i<= end2; i++)        {            if(inorder[i] == preorder[begin1])            {                break;            }        }        int leftLen = i - begin2;        root->left = buildTree(preorder,begin1+1,begin1+leftLen,inorder,begin2,begin2+leftLen-1);        root->right = buildTree(preorder,begin1+leftLen+1,end1,inorder,begin2+leftLen+1,end2);        return root;    }    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        return buildTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);    }

方法二、非递归遍历
巧用栈–可以研究研究

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        TreeNode* root = NULL;        if(preorder.size() == 0)            return root;        stack<TreeNode*> sta;        root = new TreeNode(preorder[0]);        sta.push(root);        int index=0;        for(int i = 1; i < preorder.size(); i++)        {            TreeNode* cur = sta.top();            if(sta.top()->val != inorder[index])            {                cur->left = new TreeNode(preorder[i]);                sta.push(cur->left);            }            else            {                while(!sta.empty() && sta.top()->val == inorder[index])                {                    cur = sta.top();                    sta.pop();                    index++;                }                if(index < inorder.size())                {                    cur->right = new TreeNode(preorder[i]);                    sta.push(cur->right);                }            }        }        return root;    }

二、中后序构建
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.
方法一、递归的方法
思想和上述差不多

TreeNode* buildTree(vector<int>& inorder,int begin1, int end1, vector<int>& postorder,int begin2, int end2)    {        if(begin2 > end2)            return NULL;        else if(begin2 == end2)            return new TreeNode(postorder[end2]);        TreeNode* root = new TreeNode(postorder[end2]);        int i = begin1;        for(; i<= end1; i++)        {            if(inorder[i] == postorder[end2])                break;        }        int leftLen = i - begin1;        root->left = buildTree(inorder,begin1,begin1+leftLen-1,postorder,begin2,begin2+leftLen-1);        root->right = buildTree(inorder,begin1+leftLen+1,end1,postorder,begin2+leftLen,end2-1); //注意由于end2处为根节点，所以过滤掉        return root;    }    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {        return buildTree(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);    }

方法二：非递归的实现
代码见：https://discuss.leetcode.com/topic/4746/my-comprehension-of-o-n-solution-from-hongzhi

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