轉(zhuǎn)載自：http://blog.csdn.net/sbitswc/article/details/26433051

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

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

There is an example. _______7______/ \__10__ ___2/ \ /4 3 _8\ /1 11 The preorder and inorder traversals for the binary tree above is: preorder = {7,10,4,3,1,2,8,11} inorder = {4,10,3,1,7,11,8,2}
The first node in preorder alwasy the root of the tree. We can break the tree like:
1st round:
preorder: ?{7}, {10,4,3,1}, {2,8,11}
inorder: ? ? {4,10,3,1}, {7}, {11, 8,2}

_______7______/ \{4,10,3,1} {11,8,2} Since we alreay find that {7} will be the root, and in "inorder" sert, all the data in the left of {7} will construct the left sub-tree. And the right part will construct a right sub-tree. We can the left and right part agin based on the preorder. 2nd round
left part ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?right part
preorder: {10}, {4}, {3,1} ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?{2}, {8,11}
inorder: ?{4}, {10}, {3,1} ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?{11,8}, {2}


_______7______/ \__10__ ___2/ \ /4 {3,1} {11,8} see that, {10} will be the root of left-sub-tree and {2} will be the root of right-sub-tree.
Same way to split {3,1} and {11,8}, yo will get the complete tree now.

_______7______/ \__10__ ___2/ \ /4 3 _8\ /1 11 So, simulate this process from bottom to top with recursion as following code. c++

[cpp] view plain copy

TreeNode?*BuildTreePI(??

????vector<int>?&preorder,??

????vector<int>?&inorder,??

????int?p_s,?int?p_e,??

????int?i_s,?int?i_e){??

????if(p_s?>?p_e)?return?NULL;??

????int?pivot?=?preorder[p_s];??

????int?i?=?i_s;??

????for(;i<i_e;i++){??

????????if(inorder[i]?==?pivot)??

????????????break;??

????}??

????int?length1?=?i-i_s-1;??

????int?length2?=?i_e-i-1;??

????TreeNode*?node?=?new?TreeNode(pivot);??

????node->left?=?BuildTreePI(preorder,inorder,p_s+1,length1+p_s+1,i_s,?i-1);??

????node->right?=?BuildTreePI(preorder,?inorder,?p_e-length2,?p_e,?i+1,?i_e);??

????return?node;??

}??

TreeNode?*buildTree(vector<int>?&preorder,?vector<int>?&inorder)?{??

????return?BuildTreePI(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);??

}??

java

[java] view plain copy

public?TreeNode?buildTree(int[]?preorder,?int[]?inorder)?{??

????????return?buildPI(preorder,?inorder,?0,?preorder.length-1,?0,?inorder.length-1);??

????}??

????public?TreeNode?buildPI(int[]?preorder,?int[]?inorder,?int?p_s,?int?p_e,?int?i_s,?int?i_e){??

????????if(p_s>p_e)??

????????????return?null;??

????????int?pivot?=?preorder[p_s];??

????????int?i?=?i_s;??

????????for(;i<i_e;i++){??

????????????if(inorder[i]==pivot)??

????????????????break;??

????????}??

????????TreeNode?node?=?new?TreeNode(pivot);??

????????int?lenLeft?=?i-i_s;??

????????node.left?=?buildPI(preorder,?inorder,?p_s+1,?p_s+lenLeft,?i_s,?i-1);??

????????node.right?=?buildPI(preorder,?inorder,?p_s+lenLeft+1,?p_e,?i+1,?i_e);??

????????return?node;??

????}?

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