LeetCode-in-Java

106. Construct Binary Tree from Inorder and Postorder Traversal

Medium

Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

Example 1:

Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]

Output: [3,9,20,null,null,15,7]

Example 2:

Input: inorder = [-1], postorder = [-1]

Output: [-1]

Constraints:

• 1 <= inorder.length <= 3000
• postorder.length == inorder.length
• -3000 <= inorder[i], postorder[i] <= 3000
• inorder and postorder consist of unique values.
• Each value of postorder also appears in inorder.
• inorder is guaranteed to be the inorder traversal of the tree.
• postorder is guaranteed to be the postorder traversal of the tree.

Solution

import com_github_leetcode.TreeNode;

/*
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        int[] inIndex = new int[] {inorder.length - 1};
        int[] postIndex = new int[] {postorder.length - 1};
        return helper(inorder, postorder, inIndex, postIndex, Integer.MAX_VALUE);
    }

    private TreeNode helper(int[] in, int[] post, int[] index, int[] pIndex, int target) {
        if (index[0] < 0 || in[index[0]] == target) {
            return null;
        }
        TreeNode root = new TreeNode(post[pIndex[0]--]);
        root.right = helper(in, post, index, pIndex, root.val);
        index[0]--;
        root.left = helper(in, post, index, pIndex, target);
        return root;
    }
}

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