代码随想录——从前序与中序遍历序列构造二叉树（Leetcode105）

题目链接
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递归

/*** 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.length == 0 || inorder.length == 0){return null;}return buildHelper(preorder, 0, preorder.length, inorder, 0, inorder.length);}public TreeNode buildHelper(int[] preorder, int preorderStart, int preorderEnd, int[] inorder, int inorderStart, int inorderEnd){if(preorderStart == preorderEnd){return null;}int rootVal = preorder[preorderStart];TreeNode root = new TreeNode(rootVal);int middleIndex;for(middleIndex = inorderStart; middleIndex < inorderEnd; middleIndex++){// rootvalif(inorder[middleIndex] == rootVal){break;}}// 对中序遍历处理int leftInorderStart = inorderStart; int leftInorderEnd = middleIndex;int rightInorderStart = middleIndex + 1;int rightInorderEnd = inorderEnd;// 对前序遍历处理int leftPreorderStart = preorderStart + 1; int leftPreorderEnd = preorderStart + 1 + middleIndex - inorderStart;int rightPreorderStart = preorderStart + 1 + middleIndex - inorderStart;int rightPreorderEnd = preorderEnd;root.left = buildHelper(preorder, leftPreorderStart, leftPreorderEnd , inorder, leftInorderStart, leftInorderEnd);root.right = buildHelper(preorder,rightPreorderStart, rightPreorderEnd, inorder, rightInorderStart, rightInorderEnd);return root;}
}