[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;     
    }
}