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  • 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.

    深度遍历解决,中序遍历代码如下:

    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        
        //深度遍历二叉树
        public boolean DFS(TreeNode node,int sum,int currentSum) {
            if(node==null) return false;
            if(node.left==null&&node.right==null) return (currentSum+node.val)==sum;
            else{
                return DFS(node.left,sum,currentSum+node.val)||DFS(node.right,sum,currentSum+node.val);
            }
        }
        
        public boolean hasPathSum(TreeNode root, int sum) {
            return DFS(root,sum,0);
        }
    }
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  • 原文地址:https://www.cnblogs.com/mrpod2g/p/4292856.html
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