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.

解題

Runtime: 0 ms, faster than 100%

Memory Usage: 4.1 MB, less than 63.44%

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func buildTree(inorder []int, postorder []int) *TreeNode {
    if len(inorder) == 1 {
        return &TreeNode{ Val: inorder[0] }
    }

    if len(postorder) == 0 {
        return nil
    }

    midV := postorder[len(postorder) - 1]
    for i, n := range inorder {
        if n == midV {
            left := buildTree(inorder[:i], postorder[:i])
            right := buildTree(inorder[i+1:], postorder[i:len(postorder) - 1])
            return &TreeNode{ Val: midV, Left: left, Right: right}
        }
    }

    return nil
}