leetcode Path Sum

https://oj.leetcode.com/problems/path-sum/

实际上可看做二叉树的遍历。

只不过要比二叉树的遍历多存储一个变量，就是根节点到当前节点的和。

可以深度遍历也可广度遍历

深度遍历（前序遍历）

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if(root==null)return false;          Stack<TreeNode>stack=new Stack<TreeNode>();   Stack<Integer>number=new Stack<Integer>();   stack.push(root);   number.push(root.val);   while(!stack.isEmpty()){   TreeNode temp=stack.pop();   int current=number.pop();   if(temp.right==null&&temp.left==null){ if(current==sum)return true;   }   if(temp.right!=null){   stack.push(temp.right);   number.push(temp.right.val+current);   }   if(temp.left!=null){   stack.push(temp.left);   number.push(temp.left.val+current);   }   }//while   return false;    }}

广度遍历（层次遍历）

在java中LIinkedList可做queue使用

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if(root==null)return false;           LinkedList<TreeNode>list=new LinkedList<TreeNode>();   LinkedList<Integer>number=new LinkedList<Integer>();   list.offer(root);   number.offer(root.val);   int ret=0;   while(!list.isEmpty()){   TreeNode temp=list.poll();   int cur=number.poll();   if(temp.left==null&&temp.right==null){  if(cur==sum)return true;   }   if(temp.left!=null){   list.offer(temp.left);   number.offer(temp.left.val+cur);   }   if(temp.right!=null){   list.offer(temp.right);   number.offer(temp.right.val+cur);   }   }   return false;    }}

另外说起树这种数据结构，跟它最相关的算法就是递归。

下面是一个递归算法

public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if(root==null)return false;        sum-=root.val;        if(root.left==null&&root.right==null)return sum==0;        return hasPathSum(root.left,sum)||hasPathSum(root.right,sum);    }}