94. Binary Tree Inorder Traversal（二叉树中序遍历）

 

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

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [1,3,2].

递归：

 class Solution {
     List<Integer> res = new ArrayList<Integer>();
     public List<Integer> inorderTraversal(TreeNode root) {
           help(root);
         return res;
     }
     private void help(TreeNode root){
         if(root==null) return ;
         if(root.left!=null)
             help(root.left);

         res.add(root.val);

         if(root.right!=null)
             help(root.right);
     }
 }

非递归：

终极版:

 class Solution {
     public List<Integer> inorderTraversal(TreeNode root) {
         Stack<TreeNode> s = new Stack<TreeNode>();
         List<Integer> res = new ArrayList<Integer>();
         while(root!=null||!s.isEmpty()){
             while(root!=null){
                 s.push(root);
                 root=root.left;
             }
             if(!s.isEmpty()){
                 root = s.pop();
                 res.add(root.val);
                 root = root.right;
             }
         }
         return res;
     }
 }

利用辅助桟

 class Solution {

     public List<Integer> inorderTraversal(TreeNode root) {
         List<Integer> res = new ArrayList<Integer>();
         Stack<TreeNode> stack = new Stack<TreeNode>();
         TreeNode cur = root;
         while(cur!=null ||!stack.isEmpty()){
             while(cur!=null){
                 stack.push(cur);
                 cur =cur.left;
             }
             cur = stack.pop();
             res.add(cur.val);
             cur =cur.right;
         }
         return res;
     }

 }