[LeetCode] 144. Binary Tree Preorder Traversal Java

题目：

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1\2/3

return [1,2,3].

题意及分析：给出一个二叉树，求先序遍历的结果。使用递归很简单，但是要求不用递归，所以我们可以使用一个stack来实现非递归方法，维护一个stack，保存从根节点一直到最左边的叶节点上的点，同时将这些点的值添加进结果集，达到叶节点后需要查看该节点是否有兄弟节点（即该点的父节点是否有右子节点），若有则对该兄弟节点同样遍历到最左叶节点，然后返回上一层，以此类推。具体实现看代码：

/*** 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> preorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();Stack<TreeNode> stack=new Stack<>();while(root!=null||!stack.isEmpty()){if(root!=null){stack.push(root);res.add(root.val);root = root.left;}else{root=stack.pop();root=root.right;}}return res;}
}

转载于:https://www.cnblogs.com/271934Liao/p/7017763.html

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