105.leetcode Construct Binary Tree from Preorder and Inorder Traversal(medium)[先序、中序构造二叉树]

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

对于一棵二叉树可以通过先序、中序求得这棵树的结构，也可以通过后序、中序求得这棵树的结构，下面就是采用中序、先序构造的方法，对于先序可以确定每一步的根节点，再用中序可以确定左右子树。

/** * 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* constructTree(vector<int>& preorder,int pstart,int pend, vector<int>& inorder,int istart,int iend)    {        TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode*));        root->val = preorder[pstart];        root->left = NULL;        root->right = NULL;        if(pstart == pend&&istart == iend && preorder[pstart] == inorder[istart])        {            return root; //此时没有子节点        }        int i = istart;        for(;i<=iend;i++)        {            if(inorder[i] == root->val)               break;        }        int len1 = i-istart;        if(len1>0)        {            root->left = constructTree(preorder,pstart+1,pstart+len1,inorder,istart,i-1);        }        int len2 = iend-i;        if(len2>0)        {             root->right = constructTree(preorder,pstart+1+len1,pend,inorder,i+1,iend);        }        return root;    }    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        if(preorder.size() == 0 && inorder.size() == 0) return NULL;        return constructTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);                    }};