[LeetCode-JAVA] Binary Tree Inorder Traversal
题目:
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]
.
思路:题中规定用循环代替递归求解,用辅助stack来存储遍历到的节点,如果不为空则入栈,否则出栈赋值,并指向其右节点。
代码:
public class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> req = new ArrayList<Integer>(); if(root == null) return req; Stack<TreeNode> stack = new Stack<TreeNode>(); while(!stack.isEmpty() || root != null){ if(root != null){ stack.push(root); root = root.left; }else{ TreeNode temp = stack.pop(); req.add(temp.val); root = temp.right; //最重要的一步 } } return req; } }
参考链接: http://www.programcreek.com/2012/12/leetcode-solution-of-binary-tree-inorder-traversal-in-java/