Binary Tree Inorder Traversal ——LeetCode

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree

{1,#,2,3}

,

1
\
2
/
3

return

[1,3,2]

.

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

题目大意：中序遍历一个二叉树，递归的方案太low，用迭代的方式来写？

解题思路：不用递归，那就自己实现栈呗

1、首先节点入栈，处理当前节点左孩子，并且压入栈，当左节点非空，循环遍历；

2、找到第一个左孩子为空的节点，将此节点出栈，将节点值加入结果链表，并把当前节点设为右孩子；

3、循环到栈为空。

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.add(curr);
curr = curr.left;
}
curr = stack.pop();
res.add(curr.val);
curr = curr.right;
}
return res;
}