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Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree [1,null,2,3],
1
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
求二叉树的中序遍历，要求不是用递归。
先用递归做一下，很简单。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { List result = new ArrayList<Integer>(); public List<Integer> inorderTraversal(TreeNode root) { if( root == null) return result; getResult(root); return result; } public void getResult(TreeNode root){ if( root.left != null) getResult(root.left); result.add(root.val); if( root.right != null) getResult(root.right); } }
不用递归，用栈实现也是很简单的。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { List result = new ArrayList<Integer>(); public List<Integer> inorderTraversal(TreeNode root) { if( root == null) return result; Stack stack = new Stack<TreeNode>(); TreeNode node = root; while( true ){ if( node.left == null){ result.add(node.val); if( node.right == null ){ if( stack.isEmpty() ) break; else node = (TreeNode) stack.pop(); }else{ node = node.right; } }else{ TreeNode flag = node; node = node.left; flag.left = null; stack.push(flag); } } return result; } }
leetcode 94 Binary Tree Inorder Traversal ----- java
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原文地址：http://www.cnblogs.com/xiaoba1203/p/5981983.html
