106. Construct Binary Tree from Inorder and Postorder Traversal（js）

106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   / 
  9  20
    /  
   15   7
题意：通过中序遍历和后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[]} inorder
 * @param {number[]} postorder
 * @return {TreeNode}
 */
var buildTree = function(inorder, postorder) {
            return backtrack(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
var backtrack = function( inorder,  inStart,inEnd , postorder, postStart, postEnd){
       if(inStart>inEnd || postStart>postEnd){
           return null;
       }
      let root = new TreeNode(postorder[postEnd]);
        let inIndex=0;
        for(let i=inStart;i<=inEnd;i++){
            if(inorder[i]==root.val){
                inIndex=i;
            }
        }
        root.left=backtrack(inorder,inStart,inIndex-1,postorder,postStart,postStart+inIndex-inStart-1);
        root.right=backtrack(inorder,inIndex+1,inEnd,postorder,postStart+inIndex-inStart,postEnd-1);
        return root;
    }