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Commit0d217c6

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author
applewjg
committed
ConstructBinaryTreefromInorderandPostorderTraversal
Change-Id: I8f6f38cc196de426c64029e825e1fe93c7f88722
1 parentecda904 commit0d217c6

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/*
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Author: Andy, nkuwjg@gmail.com
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Date: Jan 16, 2015
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Problem: Construct Binary Tree from Inorder and Postorder Traversal
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Difficulty: Easy
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Source: http://leetcode.com/onlinejudge#question_106
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Notes:
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Given inorder and postorder traversal of a tree, construct the binary tree.
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Note:
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You may assume that duplicates do not exist in the tree.
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Solution: Recursion.
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*/
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/**
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* Definition for binary tree
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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publicclassSolution {
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publicTreeNodebuildTree(int[]inorder,int[]postorder) {
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if(inorder.length==0||postorder.length==0||inorder.length!=postorder.length)
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returnnull;
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returnbuildTreeRe(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
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}
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publicTreeNodebuildTreeRe(int[]inorder,ints1,inte1,int[]postorder,ints2,inte2){
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if(e2<s2)returnnull;
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if(s2==e2)returnnewTreeNode(postorder[e2]);
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intj=-1;
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for(inti=s1;i<=e1;i++){
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if(inorder[i]==postorder[e2]){
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j=i;
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break;
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}
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}
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intleft_len =j-s1;
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TreeNoderoot =newTreeNode(postorder[e2]);
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root.left =buildTreeRe(inorder,s1,j-1,postorder,s2,s2+left_len-1);
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root.right =buildTreeRe(inorder,j+1,e1,postorder,s2+left_len,e2-1);
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returnroot;
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}
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}

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