【leetcode每日刷题】【!】【树/递归】124. Binary Tree Maximum Path Sum

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Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

寻找和最大的树的路径，树的路径不应该同时包括两个左叶子节点或者两个右叶子节点。使用递归的方法找到当前根节点的路径最大值，每次找到（当前根节点值、当前根节点加左右子树节点最大值、当前根节点和左子树最大值、当前根节点和右子树最大值）的最大值赋给全局变量max，每个子树的路径最大值不是当前节点和左右子树相加的值，而是（当前节点、当前节点加左子树、当前节点加右子树）的最大值，因为是路径。

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

class num124 {
    int max = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        rootSum(root);
        return max;
    }
    public int rootSum(TreeNode root){
        if(root == null) return 0;
        int ls = rootSum(root.left);
        int rs = rootSum(root.right);
        this.max = Math.max(max, root.val + ls + rs);
        this.max = Math.max(max, root.val);
        int leftAcur = root.val + ls;
        int rightAcur = root.val + rs;
        this.max = Math.max(max, leftAcur);
        this.max = Math.max(max, rightAcur);
        if(leftAcur < 0 && rightAcur < 0){
            return root.val < 0 ? 0 : root.val;
        }
        return leftAcur > rightAcur? leftAcur : rightAcur;
    }
}