106. Construct Binary Tree from Inorder and Postorder Traversal的C++解法

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使用递归：
 

class Solution {
public:
	TreeNode * buildTree(vector<int> &inorder, vector<int> &postorder) {
		// write your code here
		if (inorder.size() == 0) return NULL;
		int root = postorder[postorder.size() - 1];
		TreeNode* this_root= new TreeNode(root);
		int i;
		for (i = 0; i < inorder.size(); i++)
			if (inorder[i] == root) break;
		vector<int>new_inorder;
		vector<int>new_postorder;
		new_inorder.assign(inorder.begin(), inorder.begin() + i);
		new_postorder.assign(postorder.begin(), postorder.begin() + i);
		this_root->left = buildTree(new_inorder, new_postorder);
		new_inorder.assign(inorder.begin() + i + 1, inorder.end());
		new_postorder.assign(postorder.begin() + i, postorder.end() - 1);
		this_root->right = buildTree(new_inorder, new_postorder);
		return this_root;
	}
};

以上为了方便，直接在原方法上写的递归，所以需要多次重建两个vector，时间和空间占用得都比较多。还可以采用重写一个递归的方法：
 

class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
    }
    TreeNode *buildTree(vector<int> &inorder, int iLeft, int iRight, vector<int> &postorder, int pLeft, int pRight) {
        if (iLeft > iRight || pLeft > pRight) return NULL;
        TreeNode *cur = new TreeNode(postorder[pRight]);
        int i = 0;
        for (i = iLeft; i < inorder.size(); ++i) {
            if (inorder[i] == cur->val) break;
        }
        cur->left = buildTree(inorder, iLeft, i - 1, postorder, pLeft, pLeft + i - iLeft - 1);
        cur->right = buildTree(inorder, i + 1, iRight, postorder, pLeft + i - iLeft, pRight - 1);
        return cur;
    }
};