Description
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?
问题描述
以迭代的方式获取二叉树中序遍历的节点列表
解法
public class Solution {
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;
}
}