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.

For example, given

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

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

Analysis

关键在于Preorder中[root, left, right]的遍历顺序和Inorder的[left, root, right]顺序。因此可以从preorder中得到root在inorder中找到对应的root,于是left和right子树就可以分别进行进一步查找。

Solution

/**
 * 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[] preorder, int[] inorder) {
        if (preorder == null || inorder == null || preorder.length != inorder.length) {
            return null;
        }
       return buildTreeHelper(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1) ;
    }
    TreeNode buildTreeHelper(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) {
        if (inStart > inEnd) {
            return null;
        }
        int rootPosition = findPosition(inStart, inEnd, inorder, preorder[preStart]);
        TreeNode root = new TreeNode(preorder[preStart]);
        root.left = buildTreeHelper(preorder, inorder, preStart + 1, preStart + rootPosition - inStart, inStart, rootPosition - 1 );

        root.right = buildTreeHelper(preorder, inorder, preEnd - inEnd + rootPosition + 1, preEnd, rootPosition + 1, inEnd );
        return root;
    }
    int findPosition (int start, int end, int[] arr, int target) {
        int i;
        for (i = start; i <= end; i++) {
            if (arr[i] == target) {
                return i;
            }
        }
        return -1;
    }
}

Using Pre-built Hashmap for Index Lookup

via @yavinci: use HashMap to cache the inorder[] position.

public TreeNode buildTree(int[] preorder, int[] inorder) {
    Map<Integer, Integer> inMap = new HashMap<Integer, Integer>();

    for(int i = 0; i < inorder.length; i++) {
        inMap.put(inorder[i], i);
    }

    TreeNode root = buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inMap);
    return root;
}

public TreeNode buildTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd, Map<Integer, Integer> inMap) {
    if(preStart > preEnd || inStart > inEnd) return null;

    TreeNode root = new TreeNode(preorder[preStart]);
    int inRoot = inMap.get(root.val);
    int numsLeft = inRoot - inStart;

    root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorder, inStart, inRoot - 1, inMap);
    root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorder, inRoot + 1, inEnd, inMap);

    return root;
}

*LeetCode Official Solution

Depth First Search (DFS) - Recursion - ID Map

class Solution {
  // start from first preorder element
  int pre_idx = 0;
  int[] preorder;
  int[] inorder;
  HashMap<Integer, Integer> idx_map = new HashMap<Integer, Integer>();

  public TreeNode helper(int in_left, int in_right) {
    // if there is no elements to construct subtrees
    if (in_left == in_right)
      return null;

    // pick up pre_idx element as a root
    int root_val = preorder[pre_idx];
    TreeNode root = new TreeNode(root_val);

    // root splits inorder list
    // into left and right subtrees
    int index = idx_map.get(root_val);

    // recursion 
    pre_idx++;
    // build left subtree
    root.left = helper(in_left, index);
    // build right subtree
    root.right = helper(index + 1, in_right);
    return root;
  }

  public TreeNode buildTree(int[] preorder, int[] inorder) {
    this.preorder = preorder;
    this.inorder = inorder;

    // build a hashmap value -> its index
    int idx = 0;
    for (Integer val : inorder) {
      idx_map.put(val, idx++);
    }

    return helper(0, inorder.length);
  }
}

Reference

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/solution/

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