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/************************************************************************/
/* 43: Construct Binary Tree from Preorder and Inorder Traversal */
/************************************************************************/
/*
* Given preorder and inorder traversal of a tree, construct the binary tree.
*
* Note:
You may assume that duplicates do not exist in the tree.
* */
/*** 递归做法*********************/
/*
* 关键的地方在于：
*
* 在前序遍历的序列里，同时对左子树和右子树进行递归得到一个节点的左右节点
* */
public TreeNode buildTreeByPre_In(int[] preorder, int[] inorder) { return helper(0, 0, inorder.length - 1, preorder, inorder); } private TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) { if (preStart > preorder.length - 1 || inStart > inEnd) { return null; } TreeNode root = new TreeNode(preorder[preStart]); int inIndex = 0; // Index of current root in inorder for (int i = inStart; i <= inEnd; i++) { if (inorder[i] == root.val) { inIndex = i; break; } } root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder); root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder); return root; }
43: Construct Binary Tree from Preorder and Inorder Traversal
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原文地址：http://www.cnblogs.com/theonemars/p/4254336.html
