[LeetCode]Construct Binary Tree from Inorder and Postorder Traversal

Question

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

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本题难度Medium。

二分法

【复杂度】
时间 O(N) 空间 O(N)

【思路】
与​​[LeetCode]Construct Binary Tree from Preorder and Inorder Traversal​​相似。后序序列的特点则是根在最后，然后在根前面的那部分中：前面部分是左子树，后面的部分是右子树。因此反过来，先建立右子树，后建立左子树。

【代码】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int postEnd=0;
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        postEnd=postorder.length-1;
        return helper(0,inorder.length-1,inorder,postorder);
    }
    private TreeNode helper(int inStart,int inEnd,int[] inorder, int[] postorder){
        //base case
        if(inStart>inEnd)
            return null;

        TreeNode root=new TreeNode(postorder[postEnd--]);
        int inMid=0;
        for(int i=inStart;i<=inEnd;i++){
            if(root.val==inorder[i]){
                inMid=i;
                break;
            }
        }
        root.right=helper(inMid+1,inEnd,inorder,postorder);
        root.left=helper(inStart,inMid-1,inorder,postorder);
        return root;
    }
}