LeetCode Construct Binary Tree from Preorder and Inorder Traversal
2015-05-02 09:46
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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.
题意:二叉树前、中序构造出二叉树。
思路:经典的题目。
Note:
You may assume that duplicates do not exist in the tree.
题意:二叉树前、中序构造出二叉树。
思路:经典的题目。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { TreeNode build(int[] preorder, int[] inorder, int l, int r, int len) { TreeNode root = null; if (len < 1) return root; root = new TreeNode(preorder[l]); int index = 0; for (int i = 0; i < len; i++) if (inorder[r+i] == preorder[l]) { index = i; break; } root.left = build(preorder, inorder, l+1, r, index); root.right = build(preorder, inorder, l+1+index, r+1+index, len-1-index); return root; } public TreeNode buildTree(int[] preorder, int[] inorder) { int n = preorder.length; return build(preorder, inorder, 0, 0, n); } }
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