LeetCode: Construct Binary Tree from Inorder and Postorder Traversal

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.

地址：https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

算法：根据中序和后序构造出二叉树。细心一点应该不会错。代码：

/**
 * 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 *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.empty()) return NULL;
        return subBuildTree(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);
    }
    TreeNode *subBuildTree(vector<int> &inorder, vector<int> &postorder, int inbegin, int inend, int postbegin, int postend){
        TreeNode *root = new TreeNode(postorder[postend]);
        if(inbegin == inend && postbegin == postend){
            return root;
        }
        int key = postorder[postend];
        int i = inbegin;
        while(i <= inend && inorder[i] != key)  ++i;
        if(i > inbegin){
            root->left = subBuildTree(inorder,postorder,inbegin,i-1,postbegin,postbegin+i-1-inbegin);
        }
        if(inend > i){
            root->right = subBuildTree(inorder,postorder,i+1,inend,postbegin+i-inbegin,postend-1);
        }
        return root;
    }
};