zoukankan      html  css  js  c++  java
  • 【LeetCode】105 & 106 Construct Binary Tree from (Preorder and Inorder) || (Inorder and Postorder)Traversal

    Description:

        Given arrays recording 'Preorder and Inorder' Traversal (Problem 105) or  'Inorder and Postorder' (Problem 106), u need build the binary tree.

    Input:

        105. Preorder & Inorder traversal

        106. Inorder & Postorder traversal

    output:

        A binary tree.

    Solution:

      This solution uses the algorithm "divide and conquer". Taking an example of 105, we can get the root node from preorder travelsal then use inorder traversal to divide the range of left tree nodes and the right tree nodes. You can

    draw some simple instances on paper,  which will make you totally clear about the thinking. 

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    //105
    class Solution {
    public:
        void travel(TreeNode **root)
        {
            if(*root){
                travel(& ((*root) -> left));
                cout<<"val is "<<(*root)->val<<endl;
                travel(& ((*root) -> right));
            }
        }
    
        TreeNode* createTree(vector<int>& preorder, vector<int>& inorder, int preSta, int preEnd, int inSta, int inEnd)
        {
            if(preSta > preEnd || inSta > inEnd ) return NULL;
            TreeNode *root = new TreeNode(preorder[preSta]);
            int index;
            for(int i = inSta; i <= inEnd; i ++){
                if(inorder[i] == preorder[preSta]){
                    index = i;
                    break;
                }
            }
            root -> left = createTree(preorder, inorder, preSta + 1, preSta + index - inSta, inSta, index - 1);
            root -> right = createTree(preorder, inorder, preSta + index - inSta + 1, preEnd, index + 1, inEnd);
            return root;
        }
    
        TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
            TreeNode *root = createTree(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
            TreeNode **r = &root;
            //travel(r);
            return root;
        }
    };
    
    //106.
    class Solution {
    public:
        TreeNode* createTree(vector<int>& inorder, vector<int>& postorder, int inSta, int inEnd, int postSta, int postEnd){
            if(inSta > inEnd || postSta > postEnd)
                return NULL;
            TreeNode* root = new TreeNode(postorder[postEnd]);
            int index;
            for(int i = inEnd; i >= inSta; i --){
                if(postorder[postEnd] == inorder[i]){
                    index = i;
                    break;
                }
            }
            root -> right = createTree(inorder, postorder, index + 1, inEnd, postEnd - (inEnd - index), postEnd - 1);
            root -> left = createTree(inorder, postorder, inSta, index - 1, postSta, postSta + (index - inSta - 1));
            return root;
        }
        TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
            TreeNode* root = createTree(inorder, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1);
            return root;
        }
    };
  • 相关阅读:
    Android自注-15-Activity生命周期
    【v2.x OGE课程 15】 布局相关
    多线程——继承Thread类别
    Kivy A to Z -- Kivy 示例演示自带名单
    Android Training
    [TypeScript] Configuring a New TypeScript Project
    [TypeScript] Installing TypeScript and Running the TypeScript Compiler (tsc)
    [Redux] Fetching Data on Route Change
    [Redux] Wrapping dispatch() to Log Actions
    [Redux] Normalizing the State Shape
  • 原文地址:https://www.cnblogs.com/luntai/p/5655571.html
Copyright © 2011-2022 走看看