标签:struct binary dtree treenode tree rsa new from index
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public TreeNode buildTree(int[] preorder, int[] inorder) { 12 if (preorder == null) return null; 13 14 int preLeft = 0, preRight = preorder.length - 1, inLeft = 0, inRight = inorder.length -1; 15 return buildTree(preorder, inorder, preLeft, preRight, inLeft, inRight); 16 } 17 18 private TreeNode buildTree(int[] preorder, int[] inorder, int preLeft, int preRight, int inLeft, int inRight) { 19 if (preLeft > preRight) return null; 20 21 TreeNode temp = new TreeNode(preorder[preLeft]); 22 if (preLeft == preRight) 23 return temp; 24 25 int index = inLeft; 26 while (index < inRight && inorder[index] != preorder[preLeft]) 27 index++; 28 29 temp.left = buildTree(preorder, inorder, preLeft + 1, preLeft + index - inLeft, inLeft, index - 1); 30 temp.right = buildTree(preorder, inorder, preLeft + index - inLeft + 1, preRight, index + 1, inRight); 31 return temp; 32 } 33 }
Construct Binary Tree From Preorder and Inorder Traversal
标签:struct binary dtree treenode tree rsa new from index
原文地址:http://www.cnblogs.com/amazingzoe/p/6688582.html