# Solution For Operations On Tree

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## Step by Step Implementation For Operations On Tree

Given the root of a binary tree, return the sum of every tree node's data. int sum(TreeNode* root) { if (!root) return 0; return root->data + sum(root->left) + sum(root->right); }

# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def maxDepth(self, root: TreeNode) -> int: if root is None: return 0 # Base case else: return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. Note: A leaf is a node with no children. Example: Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its minimum depth = 2.

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).” _______3______ / \ ___5__ ___1__ / \ / \ 6 _2_ 0 8 / \ 7 4 For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Tree { public class TreeNode { public int val; public TreeNode left, right; public TreeNode(int val) { this.val = val; left = right = null; } } public class Solution { public ListpostorderTraversal(TreeNode root) { List result = new List (); if (root == null) return result; Stack stack = new Stack (); stack.Push(root); TreeNode prev = null; while (stack.Count > 0) { TreeNode curr = stack.Peek(); // go down the tree. //check if current node is leaf, if so, process it and pop stack, //otherwise, keep going down if (prev == null || prev.left == curr || prev.right == curr) { //prev == null is the situation for the root node if (curr.left != null) stack.Push(curr.left); else if (curr.right != null) stack.Push(curr.right); else { stack.Pop(); result.Add(curr.val); } } //go up the tree from left node //need to check if there is a right child //if yes, push it to stack //otherwise, process parent and pop stack else if (curr.left == prev) { if (curr.right != null) stack.Push(curr.right); else { stack.Pop(); result.Add(curr.val); } } //go up the tree from right node //after coming back from right node, process parent node and pop stack. else if (curr.right == prev) { stack.Pop(); result.Add(curr.val); } prev = curr; } return result; } } }