标签:Binary TreeNode val Preorder preorder int Tree midIndex inorder
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
通过preorder 找到root node,然后通过inorder找到root node的左边全部点和右边全部点:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { if (preorder == null || preorder.length == 0) { return null; } TreeNode root = new TreeNode (preorder[0]); int midIndex = 0; for (int i= 0; i < inorder.length; i++) { if (inorder[i] == preorder[0]) { midIndex = i; break; } } root.left = buildTree(Arrays.copyOfRange(preorder, 1, midIndex+1), Arrays.copyOfRange(inorder, 0, midIndex)); root.right = buildTree(Arrays.copyOfRange(preorder, midIndex+1, preorder.length), Arrays.copyOfRange(inorder, midIndex+1, inorder.length)); return root; } }
标签:Binary,TreeNode,val,Preorder,preorder,int,Tree,midIndex,inorder 来源: https://www.cnblogs.com/incrediblechangshuo/p/13814104.html
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