[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);
    }