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?

二叉树的非递归中序遍历

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

如果有左子节点就把左子节点放入stack 然后pop出来访问 然后访问右节点