标签:pre round span star nbsp == hat OWIN may
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]
Return the following binary tree:
3 / 9 20 / 15 7
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 class Solution { 11 public TreeNode buildTree(int[] preorder, int[] inorder) { 12 if (preorder == null || inorder == null || preorder.length == 0 || inorder.length != preorder.length) return null; 13 return buildTree(inorder, 0, preorder, 0, inorder.length); 14 } 15 16 public TreeNode buildTree(int[] inorder, int inStart, int[] preorder, int preStart, int length) { 17 if (length == 0) return null; 18 TreeNode root = new TreeNode(preorder[preStart]); 19 int index = findIndex(inorder, root.val); 20 root.left = buildTree(inorder, inStart, preorder, preStart + 1, index - inStart); 21 root.right = buildTree(inorder, index + 1, preorder, preStart + (index - inStart) + 1, length - (index - inStart) - 1); 22 return root; 23 } 24 25 public int findIndex(int[] inorder, int val) { 26 for (int i = 0; i < inorder.length; i++) { 27 if (inorder[i] == val) { 28 return i; 29 } 30 } 31 return -1; 32 } 33 }
Construct Binary Tree from Preorder and Inorder Traversal
标签:pre round span star nbsp == hat OWIN may
原文地址:https://www.cnblogs.com/beiyeqingteng/p/10468683.html