leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中后序构造BST

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

题意很简答，就是根据中序遍历和后序遍历来构造二叉搜索树BST，直接递归去做。

查询index可以使用map做查询。建议和上一道题 leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal 中前序构造BST一起学习。

代码如下：

class TreeNode {     int val;     TreeNode left;     TreeNode right;     TreeNode(int x) { val = x; }}/* * 直接递归解决 * */public class Solution {    public TreeNode buildTree(int[] inorder, int[] postorder)     {        if(inorder==null || inorder.length<=0 ||postorder==null || postorder.length<=0 || inorder.length!=postorder.length)            return null;        else            return getRootByRecursion(inorder,postorder,inorder.length-1,0,inorder.length-1);    }    //递归生成二叉树    TreeNode getRootByRecursion(int[] inorder, int[] postorder,int postEnd, int inStart, int inEnd)     {        if(inStart > inEnd)            return null;        else        {            //后续的第一个节点就是父节点            TreeNode root=new TreeNode(postorder[postEnd]);            //或许在中序中的index            int index=getIndexInOrder(inorder,root.val);            //lenLeft是左子树的元素数量            int lenLeft=(index-1)-inStart +1;            //lenRight是右子树的元素数量            int lenRight=(inEnd-inStart + 1) -lenLeft -1;            //递归生成左右子树            if(lenLeft>0)                root.left=getRootByRecursion(inorder,postorder,  postEnd-lenRight-1, inStart, index-1);            if(lenRight>0)                root.right=getRootByRecursion(inorder,postorder, postEnd-1, index+1, inEnd);            return root;        }    }    //在中序中查找下标，这里也可以使用map来实现查表    int getIndexInOrder(int[] inorder, int val)    {        for(int i=0;i<inorder.length;i++)        {            if(val==inorder[i])                return i;        }        return -1;    }}

下面是C++的做法，和上一道题一模一样，不过把前序遍历换成了后序遍历，就是递归解决

代码如下：

#include <iostream>#include <vector>#include <algorithm>using namespace std;/*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)    {        if (inorder.size() <= 0)            return NULL;        else            return getAll(postorder, inorder, postorder.size()-1, 0, inorder.size() - 1);    }    TreeNode* getAll(vector<int>& p, vector<int>& i,int pbeg,int ibeg,int iend)    {        if (ibeg > iend)            return NULL;        else        {            int index = getIndex(i, p[pbeg]);            int leftLen = (index - 1) - ibeg + 1;            int rightLen = (iend - ibeg + 1) - leftLen - 1;            TreeNode* root = new TreeNode(p[pbeg]);            root->left = getAll(p, i, pbeg -rightLen - 1, ibeg, index - 1);            root->right = getAll(p, i, pbeg - 1, index+1, iend);            return root;        }    }    int getIndex(vector<int>& a, int target)    {        for (int i = 0; i < a.size(); i++)        {            if (a[i] == target)                return i;        }        return 0;    }};