二叉樹在圖論中是這樣定義的：二叉樹是一個連通的無環圖，並且每一個頂點的度不大於3。有根二叉樹還要滿足根結點的度不大於2。有了根結點之後，每個頂點定義了唯一的父結點，和最多2個子結點。然而，沒有足夠的資訊來區分左結點和右結點。如果不考慮連通性，允許圖中有多個連通分量，這樣的結構叫做森林。

這裡，我使用javascript來寫二叉樹遍歷的三種非遞迴方式，因為樓主學的是javascript，對於C，JAVA，C++這個都不是很熟，所以就只好使用javascript代替；

前序遍歷

第一種方法：

var preorderTraversal = function(root) {
    var stack = [];
    var res = [];

    var p = root;
    if
 
(root == null)return [];

    while(stack.length!=0 || p!=null){
//Side by side to join the array, and deposited in the stack, the future need to use these root nodes  into the right sub-tree
        while(p!=null){
            stack.push(p);
            res.push(p.val);
            p = p.left;
        }
      //  When p is empty, it means that both the root and
 
 the left subtree are traversed, and the right tree goes
        if(stack.length!=0){
            p = stack.pop();
            p = p.right;
        }


    }

    return res;
};

前序遍歷第二種方法：

var preorderTraversal = function(root) {
  var result = [];
  var stack = [];
  var p = root;
  while(stack.length!=0
 
 || p != null) {
      if(p != null) {
          stack.push(p);
          result.push(p.val); // Add before going to children
          p = p.left;
      } else {
          var node = stack.pop();
          p = node.right;
      }
  }
  return result;
};

中序遍歷

第一種方法：

var inorderTraversal = function(root) {
  var stack = [];
  var res = [];
  var p = root;
  if(root == null) return [];

  while( stack.length!=0 || p!=null){

      while(p!=null){
          stack.push(p);
          p = p.left;
      }

      if(stack.length!=0){
          p= stack.pop();
          res.push(p.val);
          p = p.right;
      }
  }

  return res;
};

第二種方法：

var inorderTraversal = function(root) {
  var result = [];
  var stack = [];
  var p = root;
  while(stack.length!=0 || p != null) {
      if(p != null) {
      stack.push(p);
      p = p.left;
  } else {
      var node = stack.pop();
      result.push(node.val); // Add after all left children
      p = node.right;
  }
  }
  return result;
};

後序遍歷

第一種方法：

var postorderTraversal = function(root) {
  var Stack = [];
    var result = [];

    if(root==null)
        return [];

    Stack.push(root);
    while(Stack.length!=0)
    {
      var node= Stack.pop();
        result.push(node.val);


        if(node.left)
        Stack.push(node.left);

        if(node.right)
        Stack.push(node.right);

    }
    return result.reverse();

};

第二種方法：

var postorderTraversal = function(root) {
  var result = [];
  var stack = [];
  var p = root;
  while(stack.length!=0 || p != null) {
  if(p != null) {
  stack.push(p);
  result.unshift(p.val); // Reverse the process of preorder
  p = p.right; // Reverse the process of preorder
  } else {
  var node = stack.pop();
  p = node.left; // Reverse the process of preorder
  }
  }
  return result;

};