链接: 从中序与后序遍历序列构造二叉树

题目描述

算法分析

编程代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:

    TreeNode* _bulid(vector<int>& inorder, vector<int>& postorder,
                    int postbegin,int& orderi,int postend)
    {
        if(postbegin > postend)
        {
            return nullptr;
        }

        TreeNode* root = new TreeNode(postorder[orderi]);

        //查找父节点
        int rooti = postbegin;
        while(rooti <= postend && rooti >= 0)
        {
            if(postorder[orderi] == inorder[rooti])
            {
                --orderi;
                break;
            }
            ++rooti;
        }
        
        root->right = _bulid(inorder,postorder,rooti+1,orderi,postend);
        root->left = _bulid(inorder,postorder,postbegin,orderi,rooti-1);
        return root;
    }   

    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int i = postorder.size()-1;
        TreeNode* root = _bulid(inorder,postorder,0,i,inorder.size()-1);
        return root;
    }
};