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  • 112. Path Sum 二叉树路径的和

    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.

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    1. /**
    2. * Definition for a binary tree node.
    3. * public class TreeNode {
    4. * public int val;
    5. * public TreeNode left;
    6. * public TreeNode right;
    7. * public TreeNode(int x) { val = x; }
    8. * }
    9. */
    10. public class Solution {
    11. public bool HasPathSum(TreeNode root, int sum) {
    12. return dfs(root, sum);
    13. }
    14. private bool dfs(TreeNode root, int sum) {
    15. if (root == null)
    16. return false;
    17. if (root.left == null && root.right == null && sum == root.val)
    18. return true;
    19. return dfs(root.left, sum - root.val) || dfs(root.right, sum - root.val);
    20. }
    21. }







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  • 原文地址:https://www.cnblogs.com/xiejunzhao/p/23dfbec4fc085e39a017380bcf7beba3.html
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