105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

解题思路：给出前序和中序遍历，重建二叉树，根据前序找到跟节点，然后将中序分为两部分，分别在两部分中递归找到左右孩子节点。非递归用栈保存节点，遍历前序，每次判断中序位置是否为切分点。
一刷ac，非递归需要再刷

递归方法

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode buildTree(int[] preorder, int[] inorder) {        if(preorder == null || inorder == null) return null;        return construct(preorder, 0, inorder, 0, inorder.length-1);    }    public static TreeNode construct(int[] preorder, int pre_start, int[] inorder, int in_start, int in_end){        if(pre_start >= preorder.length || in_start > in_end) return null;        TreeNode root = new TreeNode(preorder[pre_start]);        for(int i = in_start; i <= in_end; i++){            if(preorder[pre_start] == inorder[i]){                root.left = construct(preorder, pre_start+1, inorder, in_start, i-1);                root.right = construct(preorder, pre_start+i-in_start+1, inorder, i+1, in_end);            }        }        return root;    }}

非递归

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode buildTree(int[] preorder, int[] inorder) {        if(preorder == null || inorder == null || preorder.length == 0 || inorder.length == 0) return null;        Stack<TreeNode> stack = new Stack<TreeNode>();        TreeNode root = new TreeNode(preorder[0]);        stack.push(root);        int index = 0;        for(int i = 1; i < preorder.length; i++){            TreeNode node = stack.peek();            if(node.val != inorder[index]){                TreeNode tmp = new TreeNode(preorder[i]);                node.left = tmp;                stack.push(tmp);            }else{                while(!stack.isEmpty() && stack.peek().val == inorder[index]){                    node = stack.pop();                    index++;                }                TreeNode tmp = new TreeNode(preorder[i]);                node.right = tmp;                stack.push(tmp);            }        }        return root;    }}