105. Construct Binary Tree from Preorder and Inorder Traversal

Problem

Tags: Array, Hash Table, Divide and Conquer, Tree, Binary Tree

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

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

Constraints:

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

Code

JS

// 105. Construct Binary Tree from Preorder and Inorder Traversal (6/15/53405)
// Runtime: 100 ms (72.57%) Memory: 41.86 MB (94.44%) 

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {number[]} preorder
 * @param {number[]} inorder
 * @return {TreeNode}
 */
var buildTree = function(preorder, inorder) {
    // 以便從節點 value 來反查在 inorder 中的位置
    let inorder_location = new Map();
    for (let i = 0; i < inorder.length; i++) inorder_location.set(inorder[i], i);
    
    function tree_build(preorder_root_location, inorder_left_bound, inorder_right_bound) {
        // 建構 root 節點，獲取 root 在 inorder 的位置
        let root_value = preorder[preorder_root_location],
            root = new TreeNode(root_value),
            inorder_root_location = inorder_location.get(root_value);
        
        // 以左子節點為 root 建構子樹
        if (inorder_root_location > inorder_left_bound) 
            root.left = tree_build(preorder_root_location + 1, inorder_left_bound, inorder_root_location - 1)
        // 以右子節點為 root 建構子樹
        if (inorder_root_location < inorder_right_bound) 
            root.right = tree_build(preorder_root_location + inorder_root_location - inorder_left_bound + 1, inorder_root_location + 1, inorder_right_bound)
        
        // 回傳 root，包含其子樹
        return root;
    }
    
    // 開始建構這棵樹，從 root (preorder[0]) 開始做，inorder 的範圍為整個的 inorder
    return tree_build(0, 0, inorder.length - 1);
};