[LeetCode-21]Construct Binary Tree from Preorder and Inorder Traversal
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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 roundleft 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++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
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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