[leetcode]94. Binary Tree Inorder Traversal二叉树中序遍历

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

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

题意：

二叉树中序遍历

Solution1:   Recursion

code

 class Solution {
      public List<Integer> inorderTraversal(TreeNode root) {
          List<Integer> list = new ArrayList<>();
          storeInorder(root, list);
          return list;
      }
     public void storeInorder(TreeNode node, List<Integer> list) {
         if(node == null)  return;
         storeInorder(node.left, list);
         list.add(node.val);
         storeInorder(node.right, list);
     }
 }

Solution2:   Iteration

code

 class Solution {
     public List<Integer> inorderTraversal(TreeNode root) {
         //left -- root -- right
         ArrayList<Integer> result = new ArrayList<>();
         Stack<TreeNode> s = new Stack<>();
         TreeNode p  = root;

         while(!s.isEmpty() || p!=null){
             if(p!=null){
                 s.push(p);
                 p = p.left;
             }else{
                 p =s.pop();
                 result.add(p.val);
                 p = p.right;
             }
         }
         return result;
     }
 }