105. Construct Binary Tree from Preorder and Inorder Traversal（js）

105. 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
题意：通过先序遍历和中序遍历构建二叉搜索树
代码如下：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} preorder
 * @param {number[]} inorder
 * @return {TreeNode}
 */
var buildTree = function(preorder, inorder) {
     return backtrack(0,0,preorder.length-1,preorder,inorder);
    }
var backtrack = function(preIndex,inStart,inEnd, preorder, inorder){
        if(preIndex>preorder.length-1 || inStart>inEnd){
            return null;
        }
        let root=new TreeNode(preorder[preIndex]);
        let inIndex=0;
        for(let i=inStart;i<=inEnd;i++){
            if(inorder[i]==root.val){
                inIndex=i;
            }
        }
        root.left=backtrack(preIndex+1,inStart,inIndex-1,preorder,inorder);
        root.right=backtrack(preIndex+1+inIndex-inStart,inIndex+1,inEnd,preorder,inorder);
        
        return root;
    }