Given a binary tree, return all root-to-leaf paths.

Example

Given the following binary tree:

1/ \ 2 3\5

All root-to-leaf paths are:

["1->2->5","1->3" ]
Tags? Binary Tree?Facebook?Binary Tree Traversal?Google

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分治法。注意一下葉子節點和null節點這次處理不太一樣而已。

1.如果是list<String> result; 不能直接result.add(Integer)！但result.add("" + Integer);可以。 result.add(Integer + s)也可以。不能直接塞一個其他類型，前面加空或者其他部分有加該類型是幫你說了要類型轉換。

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/*** Definition of TreeNode:* public class TreeNode {* public int val;* public TreeNode left, right;* public TreeNode(int val) {* this.val = val;* this.left = this.right = null;* }* }*/public class Solution {/** @param root: the root of the binary tree* @return: all root-to-leaf paths*/public List<String> binaryTreePaths(TreeNode root) {// write your code here List<String> result = new ArrayList<String>();if (root == null) {return result;}if (root.left == null && root.right == null) {result.add("" + root.val);return result;}List<String> left = binaryTreePaths(root.left);List<String> right = binaryTreePaths(root.right);for (String s : left) {result.add(root.val + "->" + s);}for (String s : right) {result.add(root.val + "->" + s);}return result;} }

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轉載于:https://www.cnblogs.com/jasminemzy/p/7637357.html

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