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.

class Solution {
public:
    TreeNode *buildTree(vector<int>& inorder, int in_left,int in_right,
                        vector<int>& postorder,  int post_left, int post_right){
        if(in_right < in_left || post_right < post_left) return NULL;
        TreeNode *root = new TreeNode(postorder[post_right]);
        int index = in_left;
        for( ; index <= in_right; ++ index ) if(inorder[index] == postorder[post_right])  break;
        int right_cnt = in_right-index;
        root->left = buildTree(inorder,in_left,index-1,postorder,post_left,post_right-right_cnt-1);
        root->right = buildTree(inorder,index+1,in_right,postorder, post_right-right_cnt,post_right-1);
        return root;
    }

    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.size() == 0) return NULL;
        else return buildTree(inorder,0,inorder.size()-1, postorder,0,postorder.size()-1);
    }
};
原文地址:https://www.cnblogs.com/xiongqiangcs/p/3821197.html