二叉树的中序遍历，左——> 根——>右

递归实现

• Runtime: 68 ms, faster than 90.03%
• Memory Usage: 36.8 MB, less than 35.02%

1、递归过程会调用函数自身，导致函数调用栈的频繁进栈出栈，时间复杂度高，每次函数调用，需要在函数栈中分配内存空间以保存参数、返回地址以及临时变量，空间复杂度高，并且有可能存在重复计算。
2、调用栈可能会溢出，每一次函数调用会在内存栈中分配空间，每个进程的栈的容量是有限的，当调用层次太多时，会超出栈的容量，从而导致栈溢出。

/**
* Definition for a binary tree node.
* function TreeNode(val) {
*     this.val = val;
*     this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var inorderTraversal = function(root) {
   var res = []
   inorder(root, res)
   return res
};
var inorder = function(root, res){
   if (root !== null){
       root.left && inorder(root.left, res)
       res.push(root.val)
       root.right && inorder(root.right, res)
   }
   else return
}

递归实现2

• Runtime: 72 ms, faster than 79.34%
• Memory Usage: 36.8 MB, less than 33.47%

/**
 * 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 {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function(root) {
    if(!root) return []
    let res = []
    
    let recursive = function(node) {
        if(node.left) recursive(node.left)
        res.push(node.val)
        if(node.right) recursive(node.right)
    }
    recursive(root)
    return res
};

非递归实现

• Runtime: 76 ms, faster than 60.86%
• Memory Usage: 37.2 MB, less than 5.01%
  当根节点非空时，将根节点入栈，左子节点入栈，直到为空时，该节点弹出，进入结果数组，然后根节点弹出，进入结果数组，对右子树进行遍历。

var inorderTraversal = function(root) {
   var stack = []
   var res = []
   while(root !== null || stack.length !== 0){
       if(root !== null){
           stack.push(root)
           root = root.left
       } else{
           root = stack.pop()
           res.push(root.val)
           root = root.right
       }
   }
   return res
};

/**
 * 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 {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function(root) {
    if(!root) return []
    let stack = []
    let res = []
    let p = root
    
    while(stack.length > 0 || p) {
        if(p) {
            stack.push(p)
            p = p.left
        } else {
            let node = stack.pop()
            res.push(node.val)
            p = node.right
        } 
    }
    return res
};