[Leetcode-Tree] Path Sum I II III

notebin 發(fā)布于2019-08-16 10:26 / 1131人閱讀

摘要：解題思路利用遞歸，對(duì)于每個(gè)根節(jié)點(diǎn)，只要左子樹和右子樹中有一個(gè)滿足，就返回每次訪問一個(gè)節(jié)點(diǎn)，就將該節(jié)點(diǎn)的作為新的進(jìn)行下一層的判斷。代碼解題思路本題的不同點(diǎn)是可以不從開始，不到結(jié)束。代碼當(dāng)前節(jié)點(diǎn)開始當(dāng)前節(jié)點(diǎn)左節(jié)點(diǎn)開始當(dāng)前節(jié)點(diǎn)右節(jié)點(diǎn)開始

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.解題思路
利用遞歸，對(duì)于每個(gè)根節(jié)點(diǎn)，只要左子樹和右子樹中有一個(gè)滿足，就返回true;
每次訪問一個(gè)節(jié)點(diǎn)，就將sum-該節(jié)點(diǎn)的val,作為新的Sum進(jìn)行下一層的判斷。
直到葉子節(jié)點(diǎn)，且sum與節(jié)點(diǎn)val相等，則表示存在這樣的path,返回true.
2.代碼

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.解題思路

本題是上一題的擴(kuò)展，需要列出所有滿足條件的path.我們只要在遞歸函數(shù)里添加List pre參數(shù)來存儲(chǔ)已經(jīng)生成的節(jié)點(diǎn)序列即可。

2.代碼

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.解題思路

本題的不同點(diǎn)是path可以不從root開始，不到leaf結(jié)束。但由于可以存在負(fù)數(shù)節(jié)點(diǎn)，所以沒法通過比較大小來縮進(jìn)節(jié)點(diǎn)，所以我們就只能考慮從每一個(gè)節(jié)點(diǎn)開始的情況。

2.代碼

public class Solution {
    public int pathSum(TreeNode root, int sum) {
        if(root==null) return 0;
        //helper(root,sum) 當(dāng)前節(jié)點(diǎn)開始
        //pathSum(root.left,sum) 當(dāng)前節(jié)點(diǎn)左節(jié)點(diǎn)開始
        //pathSum(root.right,sum) 當(dāng)前節(jié)點(diǎn)右節(jié)點(diǎn)開始
        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);
    }
}