LeetCode Construct Binary Tree from Inorder and Postorder Traversal

Description

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

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

The original problem is here.

The original code is here.

My Solution

I solve this problem in C++, as below:

/*
*Construct Binary Tree from Inorder and Postorder Traversal
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<vector>
#include<stdlib.h>
using namespace std;


//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>& inorder, vector<int>& postorder) {
        int len1 = postorder.size();
        int len2 = inorder.size();
        if(len1 != len2)
            return NULL;
        if(len1 <= 0)
            return NULL;
        TreeNode * root = build(inorder, 0, len1-1, postorder, 0, len1-1);
        return root;
    }
    TreeNode* build(vector<int>& inorder, int ni, int nj, vector<int>& postorder, int pi, int pj){
        if(pi > pj)
            return NULL;
        if(ni > nj)
            return NULL;
        
        int val = postorder[pj];
        TreeNode *node = new TreeNode(val);
        if(pi == pj){
            return node;
        }
        int pos = ni;
        for(int i=ni;i<=nj;i++){
            if(inorder[i] == val) {
                pos = i;
                break;
            }
        }
        int leftLen  = pos-ni;
        int rightLen = nj-pos;
        
        node->left  = build(inorder, ni, pos-1, postorder, pi, pi+leftLen-1);
        node->right = build(inorder, pos+1, nj, postorder, pj-rightLen, pj-1);
        return node;
    }
};

Note

根据后序序列，可以找到根节点，或者子树的根节点；在根据中序序列，可以分别找到左子树和右子树。递归地计算左子树和右子树，直至序列中只有一个节点，即叶子节点，就返回。