Monday, August 8, 2016

Google – Binary Tree Maximum Path Without Consecutive Nodes


Google – Binary Tree Maximum Path Without Consecutive Nodes
Leetcode 124 Binary Tree Maximum Path Sum的变形。区别就是不能有连续的node。
[Analysis]
有点像House Robber III, 但是House Robber规定Path只能从root开始,大大简化了题目难度。这道题的path可以是任意的,上下左右都可以,唯一的限制就是不能使用连续node。
[Solution]
思路其实就是dfs,和leetcode124是一样的,只不过在递归回parent的时候return两个值,一个是包含当前节点的最大值,另一个是不包含当前节点的最大值。
因为这道题的状态非常复杂,既要考虑包不包含,还需要考虑正负数,外加这类dfs递归题向来想不清楚,所以是道非常有价值的题,值得反复做。

[Note]
1. return两个值需要用个class来wrap
2. 注意和0的比较,在extend current node的时候,只有当left.without或right.without大于0的情况下才能extend, 否则即使当前节点为负数,继续extend只会让path sum更小。
3. 而当不考虑当前节点的时候,最大path sum也不一定就是left.with或者right.with. left.without和right.without一样也可以是不考虑当前节点情况下的最大path sum.
class Solution {
  int result = 0;
  public int maxPathSum(TreeNode root) {
    if (root == null) {
      return 0;
    }
 
    dfs(root);
    return result;
  }
 
  private Pair dfs(TreeNode root) {
    if (root == null) {
      return null;
    }
    if (root.left == null && root.right == null) {
      return new Pair(root.val, 0);
    }
 
    Pair left = dfs(root.left);
    Pair right = dfs(root.right);
 
    // with curr node
    int curr = root.val;
    if (left.without > 0) curr += left.without;
    if (right.without > 0) curr += right.without;
    result = Math.max(result, curr);
 
    // without curr node
    curr = 0;
    curr += Math.max(0, Math.max(left.with, left.without));
    curr += Math.max(0, Math.max(right.with, right.without));
    result = Math.max(result, curr);
 
    // generate return pair
    int withCurr = root.val;
    withCurr += Math.max(0, Math.max(left.without, right.without));
    int withoutCurr = 0;
    withoutCurr = Math.max(withoutCurr, Math.max(left.with, left.without));
    withoutCurr = Math.max(withoutCurr, Math.max(right.with, right.without));
 
    return new Pair(withCurr, withoutCurr);
  }
 
  private class Pair {
    int with;
    int without;
 
    Pair(int with, int without) {
      this.with = with;
      this.without = without;
    }
  }
}
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