LeetCode—105. Construct Binary Tree from Preorder and Inorder Traversal

題目

根據前序周遊和中序周遊建立二叉樹。注意二叉樹中沒有重複的數字。

思路及解法

根據前序周遊和中序周遊的特點，通過前序周遊清單确定根節點，在中序周遊的清單裡找出根節點，左邊就是左子樹，右邊就是右子樹，中序周遊清單确定左子樹和右子樹的長度，實際在遞歸過程中，傳遞的是左右子樹的開始和終止節點

代碼

/**
 * 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) {
        return helper(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);
    }
    public TreeNode helper(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd){
        if(preStart>preEnd || inStart>inEnd) return null;
        TreeNode node = new TreeNode(preorder[preStart]);
        int index=0;
        for(int i=inStart; i<=inEnd; i++){
            if(preorder[preStart]==inorder[i]){
                index = i;
                break;
            }
        }
        node.left=helper(preorder, inorder, preStart+1, preStart+index-inStart, inStart, index-1);
        node.right=helper(preorder, inorder, preStart+index-inStart+1, preEnd, index+1, inEnd);
        return node;
        
    }
}