leetCode: 112. Path Sum

Problem

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.

解題思路：

采用遞歸的思想，(root->sum) 意味著 (root.left->sum - root.value || root.right->sum - root.value)
如果當前節(jié)點為null則返回false，如果當前節(jié)點為葉子節(jié)點則判斷該節(jié)點值是不是等于sum剿干，否則遞歸扔字。

代碼如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        if(root.left==null&&root.right==null&&root.val==sum){
            return true;
        }
        return(hasPathSum(root.right,sum-root.val)||hasPathSum(root.left,sum-root.val));
    }
}

LeetCode 226 Invert Binary Tree

Problem:

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:This problem was inspired by this original tweet by Max Howell:Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

解題思路：

還是用遞歸的思想矩距。

代碼如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root==null){
            return null;
        }
        if(root.left==null&&root.right==null){
            return root;
        }
        root.left = invertTree(root.left);
        root.right = invertTree(root.right);
        
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        return root;
    }
}