LeetCode:Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Inorder and Postorder Traversal
Total Accepted: 57352 
Total Submissions: 195355 
Difficulty: Medium
Given inorder and postorder traversal of a tree, construct the binary tree.
Note: You may assume that duplicates do not exist in the tree.
Subscribe to see which companies asked this question
Hide Tags
 
Tree Array Depth-first Search
Hide Similar Problems
 
(M) Construct Binary Tree from Preorder and Inorder Traversal
考え方:
「後順」配列の最後のノードに基づいて、「中順」配列に「ルート」ノードindexが見つかります.indexによって「中序」をindexの左右のサブツリーに分ける.
java code:

/**
 * 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[] inorder, int[] postorder) {
        
        if(inorder == null || postorder == null || inorder.length != postorder.length) return null;
        
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i=0;i<inorder.length;i++)
            map.put(inorder[i], i);
        
        return buildTree(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1, map);
    }
    
    //      
    private TreeNode buildTree(int[] inorder,int s1, int e1, int[] postorder,int s2, int e2, Map<Integer, Integer> map) {
        
        if(s1 > e1 || s2 > e2) return null;
        TreeNode root = new TreeNode(postorder[e2]);
        
        int rootIndex = map.get(postorder[e2]); //        “ ”  
        root.left = buildTree(inorder, s1, rootIndex-1, postorder, s2, s2+rootIndex-s1-1, map);
        root.right = buildTree(inorder, rootIndex+1, e1, postorder, s2+rootIndex-s1, e2-1, map);
        return root;
    }
}