标签：return [1] called ons start cal class ref int
Given a binary tree, you need to check whether sum of all covered elements is equal to sum of all uncovered elements or not.
In a binary tree, a node is called Uncovered if it appears either on left boundary or right boundary. Rest of the nodes are called covered.
For example, consider below binary tree
In above binary tree, Covered node: 6, 4, 7 Uncovered node: 8, 3, 1, 10, 14, 13 The output for this tree should be false as sum of covered and uncovered node is not same
1 public class CoverBt { 2 public static boolean checkSum(TreeNode root) { 3 if(root == null) { 4 return true; 5 } 6 int sumAll = sum(root); 7 int sumUncovered = uncoveredSum(root); 8 return sumUncovered == (sumAll - sumUncovered); 9 } 10 private static int sum(TreeNode root) { 11 if(root == null) { 12 return 0; 13 } 14 return root.val + sum(root.left) + sum(root.right); 15 } 16 private static int uncoveredSumLeft(TreeNode node) { 17 if(node.left == null && node.right == null) { 18 return node.val; 19 } 20 if(node.left != null) { 21 return node.val + uncoveredSumLeft(node.left); 22 } 23 return node.val + uncoveredSumLeft(node.right); 24 } 25 private static int uncoveredSumRight(TreeNode node) { 26 if(node.left == null && node.right == null) { 27 return node.val; 28 } 29 if(node.right != null) { 30 return node.val + uncoveredSumRight(node.right); 31 } 32 return node.val + uncoveredSumRight(node.left); 33 } 34 private static int uncoveredSum(TreeNode root) { 35 int lb = 0, rb = 0; 36 if(root.left != null) { 37 lb = uncoveredSumLeft(root.left); 38 } 39 if(root.right != null) { 40 rb = uncoveredSumRight(root.right); 41 } 42 return root.val + lb + rb; 43 } 44 public static void main(String[] args) { 45 TreeNode[] nodes = new TreeNode[9]; 46 nodes[0] = new TreeNode(8); 47 nodes[1] = new TreeNode(3); 48 nodes[2] = new TreeNode(10); 49 nodes[3] = new TreeNode(1); 50 nodes[4] = new TreeNode(6); 51 nodes[5] = new TreeNode(14); 52 nodes[6] = new TreeNode(4); 53 nodes[7] = new TreeNode(7); 54 nodes[8] = new TreeNode(13); 55 nodes[0].left = nodes[1]; nodes[0].right = nodes[2]; 56 nodes[1].left = nodes[3]; nodes[1].right = nodes[4]; 57 nodes[2].right = nodes[5]; 58 nodes[4].left = nodes[6]; nodes[4].right = nodes[7]; 59 nodes[5].left = nodes[8]; 60 System.out.println(checkSum(nodes[0])); 61 } 62 }
For calculating sum of Uncovered nodes we will follow below steps:
1) Start from root, go to left and keep going left as long as left child is available, if not go to right child and again follow same procedure of going left until you reach a leaf node.
2) After step 1 sum of left boundary will be stored, now for right part again do the same procedure but now keep going to right as long as right child is available, if not then go to left child and follow same procedure of going right until you reach a leaf node.
After above 2 steps sum of all Uncovered node will be stored, we can subtract it from total sum and get sum of covered elements and check for equines of binary tree.
[GeeksForGeeks] Check sum of covered and uncovered nodes of binary tree
标签：return [1] called ons start cal class ref int
原文地址：http://www.cnblogs.com/lz87/p/7643662.html
