Rebuilding a binary tree

Title Description: 650) this.width=650; "src=" Http://img.baidu.com/hi/jx2/j_0016.gif "alt=" j_0016.gif "/> To enter the results of the pre-order traversal and the middle sequence traversal of a binary tree, rebuild the two-fork tree.

Assuming that the input two-fork tree has a pre-sequence traversal and the result of a middle-order traversal does not contain duplicate numbers

For example, enter the pre-sequence traversal sequence {1,2,4,7,3,5,6,8} and the middle sequence traversal sequence {4,7,2,1,5,3,8,6} to re-build the two-fork tree and return the root node of the two fork tree

650) this.width=650; "src=" Http://s5.51cto.com/wyfs02/M00/82/8B/wKioL1dZFumQ-BNWAAAspgCA7Rg822.png "title=" 1.png " alt= "Wkiol1dzfumq-bnwaaaspgca7rg822.png"/>

The nodes of the binary tree are defined as follows:

struct Binarytreenode{t _data; binarytreenode<t>* _left; binarytreenode<t>* _right; Binarytreenode (const t& x)//constructor: _data (x), _left (null), _right (null) {}};

Analysis: 650) this.width=650; "src=" Http://img.baidu.com/hi/jx2/j_0020.gif "alt=" J_0020.gif "/>

650) this.width=650; "src=" Http://s3.51cto.com/wyfs02/M01/82/8B/wKioL1dZF_LAkLkoAADSlvMBuVY204.png "title=" 2.png " alt= "Wkiol1dzf_laklkoaadslvmbuvy204.png"/>

650) this.width=650; "src=" Http://s4.51cto.com/wyfs02/M01/82/8B/wKioL1dZGBGDv_vSAACFzDRSPbw465.png "title=" 3.png " alt= "Wkiol1dzgbgdv_vsaacfzdrspbw465.png"/>

650) this.width=650; "src=" Http://s2.51cto.com/wyfs02/M02/82/8C/wKiom1dZFzSx_FETAAA6FUokplA282.png "title=" 4.png " alt= "Wkiom1dzfzsx_fetaaa6fuokpla282.png"/>

At this point, we find the left and right subtree of the pre-sequence traversal and the results of the middle sequence traversal, we can use this method to continue to reconstruct the left and the sub-tree, so we can use recursion to solve this problem 650) this.width=650; "Src=" http://img.baidu.com/ Hi/jx2/j_0028.gif "alt=" J_0028.gif "/>

binarytreenode* rebuild (const int* prev, const  Int* in, int length, int& index, int left, int right) {assert ( Prev && in);if  (index >= length) return null; Binarytreenode* newnode = new binarytreenode (Prev[index]);if  (left == right return newnode;//find the position of the pre-order node in the middle sequence sequence Int inidx = findindex (In, prev[index], left,  right); if (inidx < 0) return null; Newnode->_left = rebuild (prev, in, n, ++index, left, inidx - 1 ); Newnode->_right = rebuild (prev, in, n, ++index, inidx + 1,  right); return newnode;} 

650) this.width=650; "src=" Http://img.baidu.com/hi/jx2/j_0057.gif "alt=" J_0057.gif "/> Where findindex (const int* A, const int& x, int left, int right) to find the subscript of x in array a

int FindIndex (const int* A, const int& x, int left, int. right) {for (int i = left; I <= right; ++i) {if (x = = A[i]) re Turn I;} return-1;}

650) this.width=650; "src=" Http://img.baidu.com/hi/jx2/j_0060.gif "alt=" j_0060.gif "/> Note: parameters in the Index is a reference that is used to record the subscript of the current root node in the sequence of the preceding sequence, which should begin with 0.

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Rebuilding a binary tree