LeetCode 94. 二叉树的中序遍历

递归版：

class Solution {
    public List inorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        if (root == null) return res;
        help(root, res);
        return res;
    }
    public void help(TreeNode root, List list) {
        if (root == null) return ;

        help(root.left, list);
        list.add(root.val);
        help(root.right, list);
    }
}

非递归版：

        思路：

        用栈进行模拟，当栈非空或者root不为空的时候，就一直判断

        如果root为空，说明左孩子空了，判断pop()节点，然后走右子树；否则，一直往栈里面扔左孩子

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List inorderTraversal(TreeNode root) {
        // 非递归
        List res = new ArrayList<>();
        Stack stack = new Stack<>();
        while (stack.size() > 0 || root != null) {
            if (root == null) {
                TreeNode node = stack.pop();
                res.add(node.val);
                if (node.right != null) root = node.right;
            } else {
                stack.push(root);
                root = root.left;
            }
        }
        return res;
    }
}