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Path Sum总结

Path Sum类型题的思路就是进行DFS。编写DFS函数的时候要想明白：1.DFS函数的作用是干什么 2.DFS函数的base case是什么 3.DFS函数的输入输出分别是什么

  1. Path Sum
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/*
112. Path Sum
Easy

1198

380

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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.

Note: A leaf is a node with no children.

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.
*/
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null)
            return false;
        if(root.left==null && root.right==null && root.val==sum)
            return true;
        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
    }
}
  1. Path Sum II: 这道题的关键是判断好recursive的base case，不像其他很多的DFS的base case直接就是node==null，这里的base case应该有两个：1. node==null，直接返回 2. node不为空，且node.left==null&&node.right==null&&node.val==sum，这个时候我们把存下来的路线加入到结果中。
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/*
113. Path Sum II
Medium

1131

42

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

Note: A leaf is a node with no children.

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]
]
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> ans = new LinkedList<>();
        dfs(root, ans, sum, new LinkedList<>());
        return ans;
    }
    
    private void dfs(TreeNode node, List<List<Integer>> ans, int sum, List<Integer> l){
        if(node==null){
            return;
        }
        
        if(node.left==null && node.right==null && node.val==sum){
            l.add(node.val);
            ans.add(new LinkedList<>(l));
            l.remove(l.size()-1);
            return;
        }
        
        l.add(node.val);
        dfs(node.left, ans, sum-node.val, l);
        dfs(node.right, ans, sum-node.val, l);
        l.remove(l.size()-1);
    }
}
  1. Path Sum III: 这道题可以直接DFS做，但是要考虑到可能的path不止可以是从root开始，而是可以从任意一个node开始，所以如果直接DFS的话要用两个DFS函数，而且相当于brute force。
    比较好的方法是使用类似two sum的presum做法，使用一个map存储遇到过的presum的frequency。
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/*
437. Path Sum III
Easy

2320

144

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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
*/
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int pathSum(TreeNode root, int sum) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, 1);
        return helper(root, 0, sum, map);
    }
    
    private int helper(TreeNode node, int curSum, int sum, Map<Integer, Integer> map){
        if(node==null){
            return 0;
        }
        
        curSum = curSum + node.val;
        
        int res = map.getOrDefault(curSum-sum, 0);
        map.put(curSum, map.getOrDefault(curSum, 0)+1);
        res = res + helper(node.left, curSum, sum, map)+helper(node.right, curSum, sum, map);
        map.put(curSum, map.get(curSum)-1);
        return res;
    }
}

下面两道题的思路很类似，都是求maximum path sum，思路就是正常进行DFS，DFS求的是从input node到叶节点的max path sum，而不是直接求整棵树的max path sum，在过程中，我们使用一个max变量来计算目前遇到过的最大的pathsum。

  1. Binary Tree Maximum Path Sum： 
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/*
124. Binary Tree Maximum Path Sum
Hard

2158

169

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Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6
Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxPathSum(TreeNode root) {
        int[] ans = new int[]{Integer.MIN_VALUE};
        longestPath(ans, root);
        return ans[0];
    }
    
    private int longestPath(int[] ans, TreeNode node){
        if(node==null){
            return 0;
        }
        int left = Math.max(0, longestPath(ans, node.left));
        int right = Math.max(0, longestPath(ans, node.right));
        ans[0] = Math.max(ans[0], left+right+node.val);
        return Math.max(left, right) + node.val;
    }
}
  1. Diameter of Binary Tree
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/*
543. Diameter of Binary Tree
Easy

1780

111

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Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

Example:
Given a binary tree
          1
         / \
        2   3
       / \     
      4   5    
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int diameterOfBinaryTree(TreeNode root) {
        int[] ans = new int[]{0};
        longestPath(ans, root);
        return ans[0];
    }
    
    private int longestPath(int[] ans, TreeNode node){
        if(node==null){
            return 0;
        }
        
        int left = longestPath(ans, node.left);
        int right = longestPath(ans, node.right);
        ans[0] = Math.max(ans[0], left+right+1-1);
        return Math.max(left, right)+1;
    }
}
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