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LeetCode: Binary Tree Inorder Traversal 解题报告
Binary Tree Inorder Traversal
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?
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
SOL:
包括递归与非递归方法：
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */10 public class Solution {11 public List<Integer> inorderTraversal1(TreeNode root) {12 List<Integer> ret = new ArrayList<Integer>();13 rec(root, ret);14 return ret;15 }16 17 public void rec(TreeNode root, List<Integer> ret) {18 if (root == null) {19 return;20 }21 22 rec(root.left, ret);23 ret.add(root.val);24 rec(root.right, ret);25 }26 27 public List<Integer> inorderTraversal(TreeNode root) {28 List<Integer> ret = new ArrayList<Integer>();29 if (root == null) {30 return ret;31 }32 33 Stack<TreeNode> s = new Stack<TreeNode>();34 TreeNode cur = root;35 36 while (true) {37 while (cur != null) {38 s.push(cur);39 cur = cur.left;40 }41 42 if (s.isEmpty()) {43 break;44 }45 46 cur = s.pop();47 ret.add(cur.val);48 49 cur = cur.right;50 }51 52 return ret;53 }54 }
LeetCode: Binary Tree Inorder Traversal 解题报告
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