[Leetcode-Tree] Path Sum I II III

notebin 发布于2019-08-16 10:26 / 1090人阅读

摘要：解题思路利用递归，对于每个根节点，只要左子树和右子树中有一个满足，就返回每次访问一个节点，就将该节点的作为新的进行下一层的判断。代码解题思路本题的不同点是可以不从开始，不到结束。代码当前节点开始当前节点左节点开始当前节点右节点开始

Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

      5
     / 
    4   8
   /   / 
  11  13  4
 /        
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

1.解题思路
利用递归，对于每个根节点，只要左子树和右子树中有一个满足，就返回true;
每次访问一个节点，就将sum-该节点的val,作为新的Sum进行下一层的判断。
直到叶子节点，且sum与节点val相等，则表示存在这样的path,返回true.
2.代码

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

Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path"s sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

          5
         / 
        4   8
       /   / 
      11  13  4
     /      / 
    7    2  5   1

return
[
[5,4,11,2],
[5,8,4,5]
]

1.解题思路

本题是上一题的扩展，需要列出所有满足条件的path.我们只要在递归函数里添加List pre参数来存储已经生成的节点序列即可。

2.代码

public class Solution {
    List> res=new ArrayList>();
    public List> pathSum(TreeNode root, int sum) {
        if(root==null) return res;
        helper(root,sum,new ArrayList());
        return res;
    }
    private void helper(TreeNode root, int sum,List pre){
        if(root==null) return;
        List cur=new ArrayList(pre);
        cur.add(root.val);
        if(root.left==null&&root.right==null&&sum==root.val){
            res.add(cur);
            return;
        }
       
        helper(root.left,sum-root.val,cur);
        helper(root.right,sum-root.val,cur);
    }
}

Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  
    5   -3
   /     
  3   2   11
 /    
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

1.解题思路

本题的不同点是path可以不从root开始，不到leaf结束。但由于可以存在负数节点，所以没法通过比较大小来缩进节点，所以我们就只能考虑从每一个节点开始的情况。

2.代码

public class Solution {
    public int pathSum(TreeNode root, int sum) {
        if(root==null) return 0;
        //helper(root,sum) 当前节点开始
        //pathSum(root.left,sum) 当前节点左节点开始
        //pathSum(root.right,sum) 当前节点右节点开始
        return helper(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum);
       
    }
    private int helper(TreeNode root,int sum){
       if(root==null) return 0;
       int count=0;
       if(root.val==sum) count++;
       return count+helper(root.left,sum-root.val)+helper(root.right,sum-root.val);
    }
}

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