转载自：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;
- }
