LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal

106. Construct Binary Tree from Inorder and Postorder Traversal

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.

For example, given

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

Return the following binary tree:

  3
 / \
9  20
  /  \
 15   7

Solution

Language: Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0) {
            return null;
        }
        int len = postorder.length;
        return helper(inorder, postorder, 0, len - 1, 0, len - 1);
    }
    
    private TreeNode helper(int[] inorder, int[] postorder, int iStart, int iEnd, int pStart, int pEnd) {
        if (iStart > iEnd || pStart > pEnd) {
            return null;
        }
        int pivot = postorder[pEnd];
        TreeNode root = new TreeNode(pivot);
        int pivotIndex = index(inorder, iStart, iEnd, pivot);
        int leftLen = pivotIndex - iStart;
        root.left = helper(inorder, postorder, iStart, pivotIndex - 1, pStart, pStart + leftLen - 1);
        root.right = helper(inorder, postorder,pivotIndex + 1, iEnd, pStart + leftLen, pEnd - 1);
        return root;
    }
    
    private int index(int[] inorder, int start, int end, int target) {
        for (int i = start; i <= end; i++) {
            if (inorder[i] == target) {
                return i;
            }
        }
        return start;
    }
}

使用Map加速寻找index的过程

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0) {
            return null;
        }
        int len = postorder.length;
        Map<Integer, Integer> indexMap = new HashMap<>();
        for (int i = 0; i < len; i++) {
            indexMap.put(inorder[i], i);
        }
        return helper(inorder, postorder, 0, len - 1, 0, len - 1, indexMap);
    }
    
    private TreeNode helper(int[] inorder, int[] postorder, int iStart, int iEnd, int pStart, int pEnd, Map<Integer, Integer> map) {
        if (iStart > iEnd || pStart > pEnd) {
            return null;
        }
        int pivot = postorder[pEnd];
        TreeNode root = new TreeNode(pivot);
        int pivotIndex = map.get(pivot);
        int leftLen = pivotIndex - iStart;
        root.left = helper(inorder, postorder, iStart, pivotIndex - 1, pStart, pStart + leftLen - 1, map);
        root.right = helper(inorder, postorder,pivotIndex + 1, iEnd, pStart + leftLen, pEnd - 1, map);
        return root;
    }
}