二叉树三种遍历的非递归思路（JAVASCRIPT)

原文链接：https://blog.csdn.net/sinat_37059404/article/details/76199984
写的挺好啊。。个人觉得第二种if else的看的更清晰

二叉树在图论中是这样定义的：二叉树是一个连通的无环图，并且每一个顶点的度不大于3。有根二叉树还要满足根结点的度不大于2。有了根结点之后，每个顶点定义了唯一的父结点，和最多2个子结点。然而，没有足够的信息来区分左结点和右结点。如果不考虑连通性，允许图中有多个连通分量，这样的结构叫做森林。

这里，我使用javascript来写二叉树遍历的三种非递归方式，因为楼主学的是javascript，对于C，JAVA，C++这个都不是很熟，所以就只好使用javascript代替；

前序遍历

第一种方法：

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;
};

前序遍历第二种方法：

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;
};

中序遍历

第一种方法：

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;
};

第二种方法：

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;
};

后序遍历

第一种方法：

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();

};

第二种方法：

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;

};

</article>