剑指offer 04:重构二叉树

时间:2023-03-09 14:47:59
剑指offer 04:重构二叉树

题目描述

输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6},则重建二叉树并返回。

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
if(pre.length == 0 ||in.length == 0)
return null;
return ConstructBinaryTree(pre, 0, pre.length-1, in, 0, in.length - 1);
}
public TreeNode ConstructBinaryTree(int[] pre, int pstart, int pend, int[] in, int istart, int iend){
if(pstart > pend || istart > iend)
return null;
TreeNode root = new TreeNode(pre[pstart]);
for(int i = istart; i <= iend; i++){
if(in[i] == pre[pstart]){
root.left = ConstructBinaryTree(pre, pstart+1, pstart + i - istart, in, istart, i - 1);
root.right = ConstructBinaryTree(pre, pstart + i - istart + 1, pend, in, i + 1, iend);
break;
}
}
return root;
}
}