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  • 437. 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

    注意一下这里的路径不是从root到leaf,中间有部分的路径加和为sum也可

    time: O(n), space: O(height)

    /**
     * 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) {
            if(root == null) {
                return 0;
            }
            return dfs(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
        }
        
        public int dfs(TreeNode root, int sum) {
            if(root == null) {
                return 0;
            }
            int cnt = 0;
            if(root.val == sum) {
                cnt++;
            }
            cnt += dfs(root.left, sum - root.val);
            cnt += dfs(root.right, sum - root.val);
            return cnt;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10198793.html
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