题目描述：

根据一棵树的中序遍历与后序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

示例：

给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

返回如下的二叉树：

    3
   /   9  20
    /     15   7

解法：

# define PR pair<int, int>
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int binary_search(vector<PR>& lst, int target){
        int l = 0, r = lst.size() -1;
        int mid = 0;
        while(l <= r){
            mid = l + (r-l)/2;
            if(lst[mid].first < target){
                l = mid + 1;
            }else if(lst[mid].first == target){
                return lst[mid].second;
            }else{
                r = mid - 1;
            }
        }
        return -1;
    }
    
    // method 3: accepted
    TreeNode* buildTree(vector<int>& postorder, vector<int>& inorder, int pl, int pr, int il, int ir, vector<PR>& inlst) {
        if(pl > pr){
            return NULL;
        }else if(pl == pr){
            return new TreeNode(postorder[pr]);
        }else{
            TreeNode* root = new TreeNode(postorder[pr]);
            int mid = binary_search(inlst, postorder[pr]);
            int lsz = mid - il;
            // int rsz = ir - mid;
            root->left = buildTree(postorder, inorder, pl, pl + lsz-1, il, il + lsz-1, inlst);
            root->right = buildTree(postorder, inorder, pl+lsz, pr-1, il + lsz+1, ir, inlst);
            return root;
        }
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
       // method 3:
        int sz = postorder.size();
        int pl = 0, pr = sz-1;
        int il = 0, ir = sz-1;
        vector<PR> inlst;
        for(int i = 0; i < sz; i++){
            inlst.push_back({inorder[i], i});
        }
        sort(inlst.begin(), inlst.end());
        return buildTree(postorder, inorder, pl, pr, il, ir, inlst); 
    }
};

leetcode 106. 从中序与后序遍历序列构造二叉树(Construct Binary Tree from Inorder and Postorder Traversal)

原文：https://www.cnblogs.com/zhanzq/p/10794820.html