LeetCode-106-Construct Binary Tree from Inorder and Postorde

发布时间：2020-12-14 04:26:12 所属栏目：大数据来源：网络整理

导读：算法描述： 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 = [9,7,3] Return the following binary t

算法描述：

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 = [9,7,3]

Return the following binary tree:

    3
   /   9  20
    /     15   7

解题思路：理解原理，注意细节。

    TreeNode* buildTree(vector<int>& inorder,vector<int>& postorder) {
        return helper(inorder,0,inorder.size()-1,postorder,postorder.size()-1);
    }
    
    TreeNode* helper(vector<int>& inorder,int is,int ie,vector<int>& postorder,int ps,int pe){
        if(is > ie || ps > pe) return nullptr;
        TreeNode* node = new TreeNode(postorder[pe]);
        int pos = 0;
        for(int i=is; i <= ie; i++){
            if(postorder[pe]==inorder[i]){
                pos = i;
                break;
            }
        }
        node -> left = helper(inorder,is,pos-1,ps,ps - (is-pos+1));
        node -> right = helper(inorder,pos+1,ie,ps - (is-pos+1)+1,pe-1);
        return node;
    }

（编辑：李大同）

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