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

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

For example, given

inorder = [9,3,15,20,7]postorder = [9,15,7,20,3]

Return the following binary tree:

3   / \  9  20    /  \   15   7

题意：根据一棵树的中序遍历与后序遍历构造二叉树。可以假设树中没有重复的元素。

思路

简单的分治题目，递归解决。代码如下：

/** * 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:    TreeNode* buildTree(vector
   
    & inorder, vector
    
     & postorder) {
           return rebuild(0, inorder.size() - 1, 0, postorder.size() - 1, inorder, postorder);    }    TreeNode* rebuild(int leftin, int rightin, int leftpost, int rightpost, vector
     
      & inorder, vector
      
       & postorder) {
           if (leftin > rightin || leftpost > rightpost) return nullptr;        TreeNode *root = new TreeNode(postorder[rightpost]);        int rootin = leftin;        while (rootin <= rightin && inorder[rootin] != postorder[rightpost]) ++rootin;        int leftsize = rootin - leftin;        root->left = rebuild(leftin,  rootin - 1, leftpost, leftpost + leftsize - 1, inorder, postorder);        root->right = rebuild(rootin + 1, rightin, leftpost + leftsize, rightpost - 1, inorder, postorder);          return root;          }};
      
     
    
   

效率如下：

执行用时：52 ms, 在所有 C++ 提交中击败了26.98% 的用户内存消耗：16.9 MB, 在所有 C++ 提交中击败了85.71% 的用户

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