105. Construct Binary Tree from Preorder and Inorder Traversal

题目分析

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

根据树得前序遍历结果和中序遍历结果构造二叉树

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(inorder.length != preorder.length) {
            return null;
        }
        return myBuildTree(inorder, 0, inorder.length - 1, preorder, 0, preorder.length - 1);
    }
    
    public TreeNode myBuildTree(int[] inorder, int instart, int inend, int[] preorder, int prestart, int preend) {
        if(instart > inend) {
            return null;
        }
        // 创建根节点
        TreeNode root = new TreeNode(preorder[prestart]);
        int position = findPosition(inorder, instart, inend, preorder[prestart]);
        // position - instart 就是左子树结点数目
        root.left = myBuildTree(inorder, instart, position - 1, preorder, prestart + 1, prestart + position - instart);
        root.right = myBuildTree(inorder, position + 1, inend, preorder, position - inend + preend + 1, preend);
        return root;
    }
    
    public int findPosition(int[] arr, int start, int end, int key) {
        for(int i = start; i <= end; i++) {
            if(arr[i] == key) {
                return i;
            }
        }
        return -1;
    }
}