[Leetcode] 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?
Solution:
1. 非递归
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> inorderTraversal(TreeNode root) { 12 List<Integer> result=new ArrayList<Integer>(); 13 if(root==null) 14 return result; 15 Stack<TreeNode> s=new Stack<TreeNode>(); 16 s.add(root); 17 TreeNode p=root.left; 18 while(!s.isEmpty()){ 19 while(p!=null){ 20 s.add(p); 21 p=p.left; 22 } 23 TreeNode n=s.pop(); 24 result.add(n.val); 25 p=n.right; 26 if(p!=null){ 27 s.add(p); 28 p=p.left; 29 } 30 } 31 32 return result; 33 } 34 }
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 12 ArrayList<Integer> result = new ArrayList<Integer>(); 13 14 public ArrayList<Integer> inorderTraversal(TreeNode root) { 15 16 17 Stack<TreeNode> stack = new Stack<TreeNode>(); 18 while (root != null || !stack.empty()) { 19 while (root != null) { 20 stack.push(root); 21 root = root.left; 22 } 23 if (stack.size() > 0) { 24 root = stack.pop(); 25 result.add(root.val); 26 root = root.right; 27 } 28 } 29 30 return result; 31 32 } 33 }
2. 递归
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> inorderTraversal(TreeNode root) { 12 List<Integer> result=new ArrayList<Integer>(); 13 myInorderTraversal(root,result); 14 return result; 15 } 16 17 private void myInorderTraversal(TreeNode root, List<Integer> result) { 18 // TODO Auto-generated method stub 19 if(root!=null){ 20 myInorderTraversal(root.left, result); 21 result.add(root.val); 22 myInorderTraversal(root.right, result); 23 } 24 } 25 }