30-Day Leetcoding Challenge Day29

/**
 * 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 {
  int max_sum = Integer.MIN_VALUE;

  public int max_gain(TreeNode node) {
    if (node == null) return 0;

    // max sum on the left and right sub-trees of node
    int left_gain = Math.max(max_gain(node.left), 0);
    int right_gain = Math.max(max_gain(node.right), 0);

    // the price to start a new path where `node` is a highest node
    int price_newpath = node.val + left_gain + right_gain;

    // update max_sum if it's better to start a new path
    max_sum = Math.max(max_sum, price_newpath);

    // for recursion :
    // return the max gain if continue the same path
    return node.val + Math.max(left_gain, right_gain);
  }

  public int maxPathSum(TreeNode root) {
    max_gain(root);
    return max_sum;
  }
}

标签：node,Leetcoding,right,Day29,int,max,30,gain,TreeNode
来源： https://www.cnblogs.com/yawenw/p/12822909.html