题目描述
思路
熟练掌握迭代和递归的代码。
- 递归：额外写一个函数void inOrder(TreeNode node, List
res) - 迭代：令cur = root，一直往左子树找，找到最后一个左子节点，当cur为空，就开始处理栈顶元素(将栈顶元素加入结果集)，随后将cur设置为右子节点，继续执行以上操作。
方法一：递归
/**
* 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<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
inOrder(root, res);
return res;
}
private void inOrder(TreeNode node, List<Integer> res) {
if (node == null) return;
inOrder(node.left, res);
res.add(node.val);
inOrder(node.right, res);
}
}
方法二：迭代
/**
* 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<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
Deque<TreeNode> stack = new ArrayDeque<>();
TreeNode cur = root;
if (root == null) return res;
while (cur != null || !stack.isEmpty()) {
// 一直往左子树找,找到最后一个左子节点
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
// 当cur为空,就开始处理栈顶元素
TreeNode node = stack.pop();
res.add(node.val);
// cur设置为当前节点的右子节点
cur = node.right;
}
return res;
}
}