题解 | #重建二叉树#

/*class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param preOrder int整型一维数组 
 * @param vinOrder int整型一维数组 
 * @return TreeNode类
 */
export function reConstructBinaryTree(preOrder: number[], vinOrder: number[]): TreeNode {
    // write code here
    if(preOrder.length===0||vinOrder.length === 0)return null
    //1.找到根节点
    let rootIndex = vinOrder.indexOf(preOrder[0])
    //2.在中序排序中找到左右子树
    let leftTree = vinOrder.slice(0,rootIndex)
    let righttRee = vinOrder.slice(rootIndex+1)
    let tree = new TreeNode(preOrder[0])
    //3.再找到这个左子树根节点，右子树根节点重复上面的操作
    tree.left = reConstructBinaryTree(preOrder.slice(1,rootIndex+1),leftTree)
    tree.right = reConstructBinaryTree(preOrder.slice(rootIndex +1),righttRee)
    return tree
}