leetcode 43:construct-binary-tree-from-inorder

题目描述

给出一棵树的中序遍历和后序遍历，请构造这颗二叉树

注意：

保证给出的树中不存在重复的节点

Given inorder and postorder traversal of a tree, construct the binary tree.

Note: 
You may assume that duplicates do not exist in the tree.

示例1

输入

复制

[2,1,3],[2,3,1]

输出

复制

{1,2,3}


//不需要辅助函数，简单易懂
//后序遍历容器的最后一个数是根节点，中序遍历的根节点左边是左子树，右边是右子树，
//后序遍历左子树节点值相邻，右子树节点值也相邻。由后序遍历最后一个值将中序遍历分成
//左右两部分，再由这两部分的size将后序遍历分成左右两部分，递归即可
 
class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.empty())
            return NULL;
        int nodeval=postorder[postorder.size()-1];
        TreeNode* head=new TreeNode(nodeval);
        vector<int>::iterator iter=find(inorder.begin(),inorder.end(),nodeval);
        vector<int>leftin=vector<int>(inorder.begin(),iter);
        vector<int>rightin=vector<int>(iter+1,inorder.end());
        int left=leftin.size();
        int right=rightin.size();
        if(left>0)
        {
            vector<int>leftpo=vector<int>(postorder.begin(),postorder.begin()+left);
            head->left=buildTree(leftin,leftpo);
        }
        if(right>0)
        {
            vector<int>rightpo=vector<int>(postorder.begin()+left,postorder.begin()+left+right);
            head->right=buildTree(rightin,rightpo);
        }
        return head;
    }
};