[LeetCode] 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.

递归构造

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *createTree(vector<int> &inorder, int inBeg, int inEnd, vector<int> &postorder, int postBeg, int postEnd)
    {
        if (inBeg > inEnd)
            return NULL;
            
        int root = postorder[postEnd];
        
        int index;
        
        for(int i = inBeg; i <= inEnd; i++)
            if (inorder[i] == root)
            {
                index = i;
                break;
            }
            
        int len = index - inBeg;
        TreeNode *left = createTree(inorder, inBeg, index - 1, postorder, postBeg, postBeg + len - 1);
        TreeNode *right = createTree(inorder, index + 1, inEnd, postorder, postBeg + len, postEnd - 1);
        
        TreeNode *node = new TreeNode(root);
        node->left = left;
        node->right = right;
        
        return node;
    }
    
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (inorder.size() == 0)
            return NULL;
            
        TreeNode *head = createTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
        
        return head;
    }
};