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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?
难度：medium
题目：给定二叉树，返回其中序遍历结点。（不要使用递归）
思路：栈
Runtime: 1 ms, faster than 55.50% of Java online submissions for Binary Tree Inorder Traversal.Memory Usage: 36.2 MB, less than 100.00% of Java online submissions for Binary Tree Inorder Traversal.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) {val = x;}
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
List<Integer> result = new ArrayList<>();
while (!stack.isEmpty() || root != null) {
if (root != null) {
stack.push(root);
root = root.left;
} else {
TreeNode node = stack.pop();
result.add(node.val);
root = node.right;
}
}
return result;
}
}
