124. Binary Tree Maximum Path Sum.md

124. Binary Tree Maximum Path Sum

Question

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

Thinking:

• Method1:递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int result = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root){
        maxSum(root);
        return result;
    }
    public int maxSum(TreeNode root){
        if(root == null) return 0;
        int leftSum = maxSum(root.left);
        int rightSum = maxSum(root.right);
        int cur = root.val;
        int res = cur + ((leftSum > 0) ? leftSum:0) + ((rightSum > 0) ? rightSum:0); //包括当前节点的树的最大值。
        this.result = Math.max(result, res);
        return Math.max(cur, Math.max(cur + leftSum, cur + rightSum)); //返回到上一层的可能性有三种， 只返回当前结点，左结点往上， 右结点往上。
    }
}

Second Time

• Method 1: Recursion

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   class Solution {
   	private int max = Integer.MIN_VALUE;
   	public int maxPathSum(TreeNode root) {
   		return Math.max(max(root), this.max);
   	}
   	private int max(TreeNode node){
   		if(node == null) return 0;
   		int left = Math.max(max(node.left), 0);
   		int right = Math.max(max(node.right), 0);
   		int res = Math.max(node.val, node.val + left + right);
   		this.max = Math.max(max, res);
   		return Math.max(node.val, Math.max(node.val + left, node.val + right));
   	}
   }

Amazon Session

• Method 1: dfs + traversal 2 childs and return one

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   class Solution {
   	private int result = Integer.MIN_VALUE;
   	public int maxPathSum(TreeNode root) {
   		dfs(root);
   		return this.result;
   	}
   	private int dfs(TreeNode root){
   		if(root == null) return 0;
   		int left = Math.max(dfs(root.left), 0);
   		int right = Math.max(dfs(root.right), 0);
   		int res = Math.max(root.val, left + right + root.val);
   		this.result = Math.max(result, res);
   		return Math.max(Math.max(left, right) + root.val, root.val);
   	}
   }