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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.
类似http://www.cnblogs.com/sunshineatnoon/p/3854935.html
只是子树的前序和中序遍历序列分别更新为:
//左子树: left_prestart = prestart+1 left_preend = prestart+index-instart //右子树 right_prestart = prestart+index-instart+1 right_preend = preend
代码如下:
1 /** 2 * Definition for binary tree 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 int InorderIndex(int[] inorder,int key){ 12 if(inorder == null || inorder.length == 0) 13 return -1; 14 15 for(int i = 0;i < inorder.length;i++) 16 if(inorder[i] == key) 17 return i; 18 19 return -1; 20 } 21 public TreeNode buildTreeRec(int[] preoder,int[] inorder,int prestart,int preend,int instart,int inend){ 22 if(instart > inend) 23 return null; 24 TreeNode root = new TreeNode(preoder[prestart]); 25 int index = InorderIndex(inorder, root.val); 26 root.left = buildTreeRec(preoder, inorder, prestart+1, prestart+index-instart, instart, index-1); 27 root.right = buildTreeRec(preoder, inorder, prestart+index-instart+1, preend, index+1, inend); 28 29 return root; 30 } 31 public TreeNode buildTree(int[] preorder, int[] inorder) { 32 return buildTreeRec(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1); 33 } 34 }
【leetcode刷题笔记】Construct Binary Tree from Preorder and Inorder Traversal,布布扣,bubuko.com
【leetcode刷题笔记】Construct Binary Tree from Preorder and Inorder Traversal
标签:style blog http color strong art
原文地址:http://www.cnblogs.com/sunshineatnoon/p/3855029.html