[Leetcode][JAVA] Binary Tree Preorder Traversal, Binary Tree Inorder Traversal, Binary Tree Postorder Traversal

Binary Tree PreOrder Traversal:

Given a binary tree, return the preorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

不使用递归前序遍历，可以借助栈的帮助实现。对于一个特定节点，前序遍历顺序为: 根，左，右。所以入栈顺序为 右，左，根。由于根节点的遍历和展开（研究其右节点和左节点）是同时的，所以根节点出栈的同时即可加入遍历结果中，然后研究其右节点和左节点，如果不为空即可入栈。

 1  public List<Integer> preorderTraversal(TreeNode root) {
 2         List<Integer> ls = new ArrayList<Integer>();
 3         if(root==null)
 4             return ls;
 5         Stack<TreeNode> st = new Stack<TreeNode>();
 6         st.push(root);
 7
 8         while(!st.isEmpty())
 9         {
10             TreeNode temp = st.pop();
11             ls.add(temp.val);
12             if(temp.right!=null)
13                 st.push(temp.right);
14             if(temp.left!=null)
15                 st.push(temp.left);
16         }
17         return ls;
18     }

Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

中序遍历比前序遍历复杂一些，主要是需要区分得到的节点是需要展开还是直接遍历。一般来说第一次访问节点则展开，并且自己重新入栈，第二次从栈中访问到则计入遍历。这里采用HashSet来判断是否已经访问过。

压栈顺序为 右， 根， 左（因为中序遍历顺序为左 根 右）

 1 public List<Integer> inorderTraversal(TreeNode root) {
 2         List<Integer> ls = new ArrayList<Integer>();
 3         if(root==null)
 4             return ls;
 5         Stack<TreeNode> st = new Stack<TreeNode>();
 6         HashSet<TreeNode> hs = new HashSet<TreeNode>();
 7
 8         st.push(root);
 9         while(!st.isEmpty())
10         {
11             TreeNode temp = st.pop();
12             if(hs.contains(temp))
13             {
14                 ls.add(temp.val);
15                 continue;
16             }
17             hs.add(temp);
18             if(temp.right!=null)
19                 st.push(temp.right);
20             st.push(temp);
21             if(temp.left!=null)
22                 st.push(temp.left);
23         }
24         return ls;
25     }

Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

与中序遍历一样，只不过压栈顺序为根，右，左（后序遍历顺序为左，右，根）

 1 public List<Integer> postorderTraversal(TreeNode root) {
 2         List<Integer> ls = new ArrayList<Integer>();
 3         if(root==null)
 4             return ls;
 5         Stack<TreeNode> st = new Stack<TreeNode>();
 6         HashSet<TreeNode> hs = new HashSet<TreeNode>();
 7
 8         st.push(root);
 9         while(!st.isEmpty())
10         {
11             TreeNode temp = st.pop();
12             if(hs.contains(temp))
13             {
14                 ls.add(temp.val);
15                 continue;
16             }
17             hs.add(temp);
18             st.push(temp);
19             if(temp.right!=null)
20                 st.push(temp.right);
21             if(temp.left!=null)
22                 st.push(temp.left);
23         }
24         return ls;
25     }