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Binary Tree Inorder Traversal
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?
先贴递归实现的吧
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 List<Integer> list = new ArrayList<Integer>();12 public List<Integer> inorderTraversal(TreeNode root) {13 if(null != root){14 inorderTraversal(root.left);15 list.add(root.val);16 inorderTraversal(root.right);17 18 19 }20 return list;21 }22 }
这里要求用非递归实现，我用的堆栈
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 Stack<TreeNode> stack = new Stack<TreeNode>();14 if(null == root)15 return result;16 17 do{18 while(null != root){19 stack.push(root);20 root = root.left;21 }//左子树入栈22 TreeNode temp = stack.pop();//出栈23 result.add(temp.val);24 root = temp.right;25 if(null != root)26 {27 stack.push(root);28 root = root.left;29 }30 }while(!stack.isEmpty());31 32 return result;33 }34 }
Binary Tree Inorder Traversal
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