A tree is a hierarchy: one root, branches, and leaves, with exactly one path between any two nodes. A binary search tree (BST) adds an ordering rule: for every node, all values in the left subtree are smaller and all values in the right subtree are larger. That rule makes search, insertion and deletion fast on average, O(log n) for a balanced tree, because each comparison discards half the remaining tree. Inorder traversal of a BST (left, node, right) visits values in sorted order, which is a favourite exam fact and a quick way to check whether a tree is a BST.
Deletion has three cases. A leaf is simply removed. A node with one child is replaced by that child. A node with two children is replaced by its inorder successor (the smallest value in its right subtree) or its inorder predecessor, and that successor is then deleted from its old position. The replacement keeps the ordering rule intact, which is the whole point: any value that breaks the rule breaks the tree.
Worked example — building and reading a BST. Insert 50, 30, 70, 20, 40, 60, 80 in order. 50 is the root; 30 goes left, 70 right; 20 left of 30, 40 right of 30; 60 left of 70, 80 right of 70. Inorder traversal gives 20, 30, 40, 50, 60, 70, 80 — sorted, as it must be. Searching for 60 compares 50, then 70, then 60: three comparisons instead of scanning all seven values. If insertions arrive sorted (20, 30, 40, 50...), the tree degenerates into a chain and search becomes O(n). Balanced trees exist to prevent exactly that.