94. Binary Tree Inorder Traversal

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].

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

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Recursive:

public IList<int> InorderTraversal(TreeNode root) {
        var res = new List<int>();
        Recursive(root, res);
        return res;
        
    }
    public void Recursive(TreeNode root, IList<int> list)
    {
        if(root == null) return;
        if(root.left == null && root.right == null)
        {
            list.Add(root.val);
            return;
        }
        Recursive(root.left, list);
        list.Add(root.val);
        Recursive(root.right, list);
    }

Iteration:

此题的iteration的条件是最难把握的，因为如果每次stack push 左子树到底的时候，我们需要让loop的变量tree element指向该左元素的右子数。右子数可能为null， 此时就会跳出循环。解决方法是在循环条件里加上stack 不为空。只要不为空，则证明还有值需要iteration。

 public IList<int> InorderTraversal(TreeNode root) {
        var res = new List<int>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        while(root != null || stack.Count() >0)
        {
            while(root != null)
            {
                stack.Push(root);
                root = root.left;
            }
            var a = stack.Pop();
            res.Add(a.val);
            root = a.right;//cruz we have stack.Count() > 0 , no worry right is null
        }
        return res;
    }