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  • 437. Path Sum III(统计路径和等于一个数的路径数量)

    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

    路径不一定以 root 开头,也不一定以 leaf 结尾,但是必须连续。
    方法一:递归
    时间复杂度:o(n) 空间复杂度:o(1)
    /**
     * 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;
            int ret=0;
         //累加,每次从根节点开始迭代,找到了就返回1,找到叶子结点还找不到,就返回0。下次从根节点的叶子结点找,依次递归遍历所有可能的组合。 ret
    =pathSumStartwithRoot(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum); return ret; } private int pathSumStartwithRoot(TreeNode root, int sum){ //这个地方思路和112题类似 if(root==null) return 0; int ret=0; if(root.val==sum) ret++; ret+=pathSumStartwithRoot(root.left,sum-root.val)+pathSumStartwithRoot(root.right,sum-root.val); return ret; } }
    苟有恒,何必三更眠五更起;最无益,莫过一日暴十日寒。
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  • 原文地址:https://www.cnblogs.com/shaer/p/10570744.html
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