LeetCode105. 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.

這種類型的題目經常考選擇題。對於這道題，採用的是分治的思想，通過preorder的第一個元素爲根元素，將inorder分成兩部分，也將preorder分成兩部分。這樣每個部分又是相同的子問題。解決完這兩個子問題，就是得到了當前結點的左右子樹，將其合併即可。

public class Solution {
    private TreeNode partitionalMerge(int[] preorder, int pS, int pE, int[] inorder ,int iS, int iE){
        if(pS >pE || iS > iE) return null;
        if(pS == pE) return new TreeNode(preorder[pS]);
        int mid = 0;
        for(mid =iS; mid<=iE; mid++){
            if(inorder[mid] == preorder[pS])break;
        }
                
        TreeNode left = partitionalMerge(preorder, pS+1, pS+mid-iS, inorder, iS, mid-1);
        TreeNode right = partitionalMerge(preorder, pS+mid-iS+1, pE, inorder, mid+1, iE);
        TreeNode root = new TreeNode(preorder[pS]);
        root.left = left;
        root.right = right;
        return root;
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length != inorder.length) return null;//exception
        if(preorder == null) return null;
        return partitionalMerge(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1);
    }
}

難度是索引座標經常算錯。這道題的最壞時間複雜度爲O（NLogN），空間複雜度爲O（LogN+N），其中N爲數組長度， LogN爲遞歸佔用空間，N爲樹所用的節點空間。

時間複雜度中的N爲搜索的最壞時間，那麼如果我們將搜索算法修改讓其時間複雜度爲O（1），這樣整體的時間複雜度爲O(N)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode partitionalMerge(int[] preorder, int pS, int pE ,int iS, int iE, HashMap<Integer, Integer> map){
        if(pS >pE || iS > iE) return null;
        if(pS == pE) return new TreeNode(preorder[pS]);
        int mid = 0;
        mid =map.get(preorder[pS]);
                
        TreeNode left = partitionalMerge(preorder, pS+1, pS+mid-iS, iS, mid-1, map);
        TreeNode right = partitionalMerge(preorder, pS+mid-iS+1, pE, mid+1, iE, map);
        TreeNode root = new TreeNode(preorder[pS]);
        root.left = left;
        root.right = right;
        return root;
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length != inorder.length) return null;//exception
        if(preorder == null) return null;
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i<inorder.length; i++){
            map.put(inorder[i], i);
        }
        return partitionalMerge(preorder, 0, preorder.length-1, 0, inorder.length-1, map);
    }
}

計算時間從20多ms降低到了6ms。