112. Path Sum 112.路径总和

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.

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.

 

复习时还不会的地方：忘记这道题独有的逆向思维了，哈哈

一般的二叉树怎么遍历来着，就是DC吧

sum = sum + left或者sum + right
下次用的是什么呢？一直加左边也不对啊

逆向思维，好吧。用的是traverse，就类似于相同二叉树的那种写法

 

//conquer的具体实现：左右都是空，只检查节点本身。
        if (root.left == null && root.right == null) 
            return (root.val == sum);

 因为是在左边递归，所以参数应该是左节点root.left 

 

只有左右节点都为空的时候，才需要考虑root.val == sum这样根本性的问题
最后递归的时候，值是减去的root.val，而且左右两边有一条路径符合即可

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        //cc
        if (root == null) return false;
        
        //conquer的具体实现：左右都是空，只检查节点本身。
        if (root.left == null && root.right == null) 
            return (root.val == sum);
        
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}