[LeetCode-20]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.

the idea is similar with the former one [Leetcode-21]

java

public TreeNode buildTree(int[] inorder, int[] postorder) {return buildIP(inorder, postorder, 0, inorder.length-1, 0, postorder.length-1);    }public TreeNode buildIP(int[] inorder, int[] postorder, int i_s, int i_e, int p_s, int p_e){if(p_s>p_e)return null;int pivot = postorder[p_e];int i = i_s;for(;i<=i_e;i++){if(inorder[i]==pivot)break;}TreeNode node = new TreeNode(pivot);int lenRight = i_e-i;node.left = buildIP(inorder, postorder, i_s, i-1, p_s, p_e-lenRight-1);node.right = buildIP(inorder, postorder, i+1, i_e, p_e-lenRight, p_e-1);return node;}

c++

TreeNode *BuildTreeIP(        vector<int> &inorder,        vector<int> &postorder,        int i_s, int i_e,        int p_s, int p_e){        if(i_s > i_e) return NULL;        int pivot = postorder[p_e];        int i = i_s;        for(;i<i_e;i++){            if(inorder[i] == pivot)                break;        }        int length1 = i-i_s;        int length2 = i_e-i;        TreeNode *node = new TreeNode(pivot);        node->left = BuildTreeIP(inorder, postorder, i_s, i-1, p_s, p_s+length1-1);        node->right = BuildTreeIP(inorder, postorder, i+1, i_e, p_e-length2, p_e-1);        return node;            }    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {        return BuildTreeIP(inorder, postorder, 0, inorder.size()-1, 0, postorder.size()-1);    }