Sum Root to Leaf Numbers
Sum Root to Leaf Numbers
Given a binary tree containing digits from 0-9
only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3
which represents the number 123
.
Find the total sum of all root-to-leaf numbers.
For example,
1 / \ 2 3
The root-to-leaf path 1->2
represents the number 12
.
The root-to-leaf path 1->3
represents the number 13
.
Return the sum = 12 + 13 = 25
.
这道题,我是在先序遍历的基础上改的,添加了一个类似堆栈的list,如果是非叶子节点将其“入栈”,如果是叶子节点node“弹栈”直到node的父亲节点出现,扫描“栈底”到“栈顶”所有元素组成的数字,计算sum。直到前序遍历结束
1 import java.util.ArrayList; 2 import java.util.List; 3 4 5 6 7 class TreeNode { 8 int val; 9 TreeNode left; 10 TreeNode right; 11 TreeNode(int x) { val = x; } 12 } 13 14 public class Solution { 15 16 List<TreeNode> list = new ArrayList<TreeNode>(); 17 18 int sum = 0; 19 20 public int sumNumbers(TreeNode root) { 21 if(null == root) //空树返回0 22 return 0; 23 else if(null == root.left && null == root.right){//只有根节点返回根节点的值 24 return root.val; 25 }else{ 26 preOrder(root); //至少有一个子树 27 // list.add(root); //根节点入栈 28 } 29 30 return sum; 31 } 32 33 /** 34 * 前序遍历树 35 * @param root 36 */ 37 public void preOrder(TreeNode root){ 38 if(root == null) 39 return; 40 if(null == root.left && root.right == null){//叶子节点 41 while(list.size() != 0 && root != list.get(list.size() - 1).left && root != list.get(list.size() - 1).right){ 42 list.remove(list.size() - 1); 43 } 44 int temp = 0; 45 for(int i = 0; i < list.size(); i++){ //计算一个数 46 temp = temp * 10 + list.get(i).val; 47 } 48 temp = temp * 10 + root.val; //叶子节点加上去 49 //System.out.println("temp = " + temp); 50 sum += temp; //计算和 51 if(root == list.get(list.size() - 1).right){ //如果叶子节点是右子树,出栈 52 list.remove(list.size() - 1); 53 } 54 } 55 else{ //非叶子节点 56 while(list.size() != 0 && root != list.get(list.size() - 1).left && root != list.get(list.size() - 1).right){ 57 list.remove(list.size() - 1); 58 } //root应该与栈顶元素相同 59 list.add(root); 60 preOrder(root.left); 61 preOrder(root.right); 62 } 63 } 64 }
题目提示的是用DFS,一会儿百度一下~
看了一下discuss里面的,这个写的挺漂亮的
public class Solution { public int sumNumbers(TreeNode root) { return preOrder(root, 0); } public int preOrder(TreeNode root, int val){ if(root == null) return 0; val = val * 10 + root.val; if(null == root.left && null == root.right) return val; else return preOrder(root.left, val) + preOrder(root.right, val); } }
Please call me JiangYouDang!