update

Author: OneCodeMonkey

leetcode_solved/[editing]leetcode_0124_Binary_Tree_Maximum_Path_Sum.cpp

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+/**
+ * AC: T:O(logn) S:O(logn) (递归栈调用消耗)
+ * Runtime: 0 ms, faster than 100.00% of Java online submissions for Binary Tree Maximum Path Sum.
+ * Memory Usage: 40.5 MB, less than 92.68% of Java online submissions for Binary Tree Maximum Path Sum.
+ *
+ * 思路：树的原始递归,重点在于理解题意,一个子树要么直接贡献一条最大和路径,要么共享 root+max(左子树递归,右子树递归).用一个外部变量记录中间可能产生的最大值.
+ */
+/**
+ * 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 {
+	public int maxPathSum(TreeNode root) {
+		int leftMax = maxPathSum1(root.left);
+		int rightMax = maxPathSum1(rootright);
+
+		return
+	}
+    public int maxPathSum1(TreeNode root) {
+        if (root == null) {
+        	return 0;
+        }
+        if (root.left == null && root.right == null) {
+        	return root.val;
+        }
+        int left = maxPathSum(root.left);
+        int right = maxPathSum(root.right);
+        return root.val + (left > right ? left : right);
+    }
+}