Construct Binary Tree from Preorder and Inorder Traversal - Leetcode

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode buildTree(int[] preorder, int[] inorder) {        return buildTreeHelper(preorder, 0 ,preorder.length-1, inorder, 0, inorder.length-1);    }        private TreeNode buildTreeHelper(int[] pre, int pre_start, int pre_end, int[] in, int in_start, int in_end){        if(in_start > in_end || pre_start > pre_end)           return null;        int root_val= pre[pre_start];        int root_id=0;        for(int i=in_start; i<=in_end; i++){            if(in[i] == root_val){               root_id = i;               break;            }        }                int distance = root_id-in_start;        TreeNode root = new TreeNode(root_val);                root.left = buildTreeHelper(pre, pre_start+1, pre_start+distance, in, in_start, root_id-1);        root.right = buildTreeHelper(pre, pre_start+distance+1, pre_end, in, root_id+1, in_end);                return root;    }    }

1）找到根节点，2）求出左子树的个数，逐个深入扩展左根节点，右根节点。

*题目中的特征是：

先序遍历：<span style="background-color: rgb(255, 102, 102);">中</span>左右

中序遍历：左<span style="background-color: rgb(255, 102, 102);">中</span>右

后序遍历：左右<span style="background-color: rgb(255, 102, 102);">中</span>