Construct a binary tree based on preorder and inorder traversal sequences
Problem Description
Given two integer arrays preorder and inorder, where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct a binary tree and return its root node.
1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder 和 inorder 均 无重复 元素
inorder 均出现在 preorder
preorder 保证 为二叉树的前序遍历序列
inorder 保证 为二叉树的中序遍历序列。
example
Original title OJ link
https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
answer code
/**
* 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) {
int inbegin = 0;
int inend = inorder.length-1;
return buildTreeChild(preorder,inorder,inbegin,inend);
}
public int i = 0;
public TreeNode buildTreeChild(int[] preorder,int[] inorder,int inbegin,int inend){
if(inbegin>inend){
return null;
}
TreeNode root = new TreeNode(preorder[i]);
int rootIndex = findIndex(inorder,inbegin,inend,preorder[i]);
i++;
root.left = buildTreeChild(preorder,inorder,inbegin,rootIndex-1);
root.right = buildTreeChild(preorder,inorder,rootIndex+1,inend);
return root;
}
private int findIndex(int[] inorder,int inbegin,int inend,int k){
int j = inbegin;
for(j = inbegin;j<= inend;j++){
if(inorder[j] == k){
return j;
}
}
return -1;
}
}