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| 1 | +/* |
| 2 | +Given a binary tree, return the inorder traversal of its nodes’ values. |
| 3 | +
|
| 4 | +Example : |
| 5 | +Given binary tree |
| 6 | +
|
| 7 | +1 |
| 8 | +\ |
| 9 | +2 |
| 10 | +/ |
| 11 | +3 |
| 12 | +return [1,3,2]. |
| 13 | +
|
| 14 | +*/ |
| 15 | + |
| 16 | +/** |
| 17 | +* Definition for binary tree |
| 18 | +* class TreeNode { |
| 19 | +* int val; |
| 20 | +* TreeNode left; |
| 21 | +* TreeNode right; |
| 22 | +* TreeNode(int x) { |
| 23 | +* val = x; |
| 24 | +* left=null; |
| 25 | +* right=null; |
| 26 | +* } |
| 27 | +* } |
| 28 | +*/ |
| 29 | +public class Solution { |
| 30 | +public ArrayList<Integer> inorderTraversal(TreeNode root) { |
| 31 | +ArrayList<Integer> result = new ArrayList<Integer>(); |
| 32 | +if(root == null) |
| 33 | +return result; |
| 34 | +Stack<TreeNode> stk = new Stack<TreeNode>(); |
| 35 | + |
| 36 | +while(root != null || !stk.isEmpty() ) |
| 37 | +{ |
| 38 | +while(root != null) |
| 39 | +{ |
| 40 | +stk.push(root); |
| 41 | +root = root.left; |
| 42 | +} |
| 43 | +root = stk.pop(); |
| 44 | +result.add(root.val); |
| 45 | +root = root.right; |
| 46 | +} |
| 47 | +return result; |
| 48 | +} |
| 49 | +} |
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