105. 从前序与中序遍历序列构造二叉树

1.题目

105. 从前序与中序遍历序列构造二叉树

难度中等1341

给定一棵树的前序遍历 preorder 与中序遍历 inorder。请构造二叉树并返回其根节点。

示例 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7]
示例 2:
Input: preorder = [-1], inorder = [-1] Output: [-1]

提示:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder 和 inorder 均无重复元素
• inorder 均出现在 preorder
• preorder 保证为二叉树的前序遍历序列
• inorder 保证为二叉树的中序遍历序列

  2.题解

解题思路

1.了解递归的思路，需要创建一个树，其实前序的头就是root，然后将中序中root所在的位置找到
2.递归的分割出，左子树的preOrder和InOrder和右子树的PreOrder和InOrder
3.通过画图得出，左子树的长度为PIndex

代码

/**
 * 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) {
    return builderTreeFunc(preorder, inorder);;
};
var builderTreeFunc = function(preorder, inorder) {
    if(preorder.length === 0) return null;
    let root = new TreeNode(preorder[0]);
    let pIndex = inorder.indexOf(root.val);
    // let leftTreeNum = pIndex;
    let leftPreOrder = preorder.slice(1,pIndex+1);
    let leftInOrder = inorder.slice(0, pIndex);
    let rightPreOrder = preorder.slice(pIndex+1);
    let rightInOrder = inorder.slice(pIndex+1);
    root.left = builderTreeFunc(leftPreOrder,leftInOrder)
    root.right = builderTreeFunc(rightPreOrder,rightInOrder)
    return root;
}