• 【Construct Binary Tree from Preorder and Inorder Traversal】cpp


    题目:

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

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

    代码:

    /**
     * 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<int>& preorder, vector<int>& inorder) {
            if ( preorder.size()==0 || inorder.size()==0 ) return NULL;
            return Solution::buildTreePI(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
        }
            static TreeNode* buildTreePI(
                vector<int>& preorder, 
                int beginP, int endP, 
                vector<int>& inorder, 
                int beginI, int endI)
            {
                // terminal condition & corner case
                if ( beginP>endP ) return NULL;
                // resurisve process
                TreeNode *root = new TreeNode(-1);
                root->val = preorder[beginP];
                // find the root node in inorder traversal
                int rootPosInorder = beginI;
                for ( int i = beginI; i <= endI; ++i )
                {
                    if ( inorder[i]==root->val ) { rootPosInorder=i; break;}
                }
                int leftSize = rootPosInorder - beginI;
                int rightSize = endI - rootPosInorder;
                root->left = Solution::buildTreePI(preorder, beginP+1, beginP+leftSize, inorder, beginI, rootPosInorder-1);
                root->right = Solution::buildTreePI(preorder, endP-rightSize+1, endP, inorder, rootPosInorder+1, endI);
                return root;
            }
    };

    tips:

    经典题目。直接学习高手代码

    http://fisherlei.blogspot.sg/2013/01/leetcode-construct-binary-tree-from.html

    http://bangbingsyb.blogspot.sg/2014/11/leetcode-construct-binary-tree-from.html

    这里注意在递归传递vector元素下标的时候,一定是绝对下标(一开始疏忽写成了相对下标,debug了不少时间)

    =========================================

    第二次过这道题,大体思路还记得,顺着思路摸了下来,代码改了一次AC了。注意每次要找到的是preorder[bp]而不是preorder[0]。

    /**
     * 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<int>& preorder, vector<int>& inorder)
            {
                return Solution::build(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
            }
            TreeNode* build(
                vector<int>& preorder, int bp, int ep,
                vector<int>& inorder, int bi, int ei
                )
            {
                if ( bp>ep || bi>ei ) return NULL;
                TreeNode* root = new TreeNode(preorder[bp]);
                int pos = Solution::findPos(inorder, bi, ei, preorder[bp]);
                int left_range = pos - bi;
                root->left = Solution::build(preorder, bp+1, bp+left_range, inorder, bi, bi+left_range-1);
                root->right = Solution::build(preorder, bp+left_range+1, ep, inorder, bi+left_range+1, ei);
                return root;
            }
            int findPos(vector<int>& order, int begin, int end, int val)
            {
                for ( int i=begin; i<=end; ++i ) if (order[i]==val) return i;
            }
    };
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  • 原文地址:https://www.cnblogs.com/xbf9xbf/p/4507542.html
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