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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?
Inorder traversal for the given figure is 4 2 5 1 3.
两种方法：1）recursive 2) iterative, 用stack
Java code:
1. recursive
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> result = new ArrayList<Integer>(); if(root != null){ result.addAll(inorderTraversal(root.left)); result.add(root.val); result.addAll(inorderTraversal(root.right)); } return result; } }
2. iterative
public class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> result = new LinkedList<Integer>(); Stack<TreeNode> s = new Stack<TreeNode>(); while(!s.isEmpty() || root != null){ if(root != null){ s.push(root); root = root.left; }else{ root = s.pop(); result.add(root.val); root = root.right; } } return result; } }
Reference:
1. http://www.geeksforgeeks.org/618/
2. https://leetcode.com/discuss/64937/inorder-traversal-java-solution-both-iteration-recursion
Leetcode Binary Tree Inorder Traversal
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原文地址：http://www.cnblogs.com/anne-vista/p/4899720.html
