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A program to check if a binary tree is BST or not
A binary search tree (BST) is a node based binary tree data structure which has the following properties. • The left subtree of a node contains only nodes with keys less than the node’s key. • The right subtree of a node contains only nodes with keys greater than the node’s key. • Both the left and right subtrees must also be binary search trees. From the above properties it naturally follows that:

• Each node (item in the tree) has a distinct key.

METHOD 1 (Simple but Wrong) Following is a simple program. For each node, check if left node of it is smaller than the node and right node of it is greater than the node. itiBTsrc nd*nd) n sS(tut oe oe { i (oe= NL) f nd = UL rtr 1 eun ; / flei lf i >ta nd * * as f et s hn oe / i (oe>et! NL & nd-lf-dt >nd-dt) f nd-lf = UL & oe>et>aa oe>aa rtr 0 eun ; / flei rgti ih ! NL & nd-rgt>aaih-dt oe>aa rtr 0 eun ; / flei,rcriey telf o rgti ntaBT* * as f eusvl, h et r ih s o S / i (iBTnd-lf)| !sS(oe>ih) f !sS(oe>et | iBTnd-rgt) rtr 0 eun ; / psigalta,i' aBT* * asn l ht ts S / rtr 1 eun ;

}

This approach is wrong as this will return true for below binary tree (and below tree is not a BST because 4 is in left subtree of 3)

METHOD 2 (Correct but not efficient) For each node, check if max value in left subtree is smaller than the node and min value in right subtree greater than the node. / Rtrstu i abnr te i abnr sac te * * eun re f iay re s iay erh re / itiBTsrc nd*nd) n sS(tut oe oe { i (oe= NL) f nd = UL rtr(re; euntu) / flei temxo telf i >ta u * * as f h a f h et s hn s / i (oe>et=UL& mxau(oe>et >nd-dt) f nd-lf!NL & aVlend-lf) oe>aa rtr(as) eunfle; / flei temno tergti < ta u * * as f h i f h ih s = hn s / i (oe>ih!NL & mnau(oe>ih)aa rtr(as) eunfle; / flei,rcriey telf o rgti ntaBT* * as f eusvl, h et r ih s o S / i (iBTnd-lf)| !sS(oe>ih) f !sS(oe>et | iBTnd-rgt) rtr(as) eunfle; / psigalta,i' aBT* * asn l ht ts S / rtr(re; euntu)

}

It is assumed that you have helper functions minValue() and maxValue() that return the min or max int value from a non-empty tree

METHOD 3 (Correct and Efficient) Method 2 above runs slowly since it traverses over some parts of the tree many times. A better solution looks at each node only once. The trick is to write a utility helper function isBSTUtil(struct node* node, int min, int max) that traverses down the tree keeping track of the narrowing min and max allowed values as it goes, looking at each node only once. The initial values for min and max should be INT_MIN and INT_MAX — they narrow from there.
/ Rtrstu i tegvnte i abnr sac te * eun re f h ie re s iay erh re (fiin vrin.* efcet eso) / itiBTsrc nd*nd) n sS(tut oe oe { rtr(sSUi(oe ITMN ITMX) euniBTtlnd, N_I, N_A); } / Rtrstu i tegvnte i aBTadis * eun re f h ie re s S n t

vle ae> mnad< mx * aus r = i n = a. / itiBTtlsrc nd*nd,itmn itmx n sSUi(tut oe oe n i, n a)

Implementation: #nld icue sdoh #nld #nld / Abnr te nd hsdt,pitrt lf cid * iay re oe a aa one o et hl adapitrt rgtcid* n one o ih hl / src nd tut oe { itdt; n aa src nd*lf; tut oe et src nd*rgt tut oe ih; } ; itiBTtlsrc nd*nd,itmn itmx; n sSUi(tut oe oe n i, n a) / Rtrstu i tegvnte i abnr sac te * eun re f h ie re s iay erh re (fiin vrin.* efcet eso) / itiBTsrc nd*nd) n sS(tut oe oe { rtr(sSUi(oe ITMN ITMX) euniBTtlnd, N_I, N_A); } / Rtrstu i tegvnte i aBTadis * eun re f h ie re s S n t vle ae> mnad< mx * aus r = i n = a. / itiBTtlsrc nd*nd,itmn itmx n sSUi(tut oe oe n i, n a) { / a epyte i BT* * n mt re s S / i (oe=UL f nd=NL) rtr 1 eun ; / flei ti nd voae temnmxcntan * * as f hs oe ilts h i/a osrit / i (oe>aamx f nd-dt i | oe>aa a) rtr 0 eun ; / ohriecektesbre rcriey * tews hc h utes eusvl, tgtnn temno mxcntan * iheig h i r a osrit / rtr eun iBTtlnd-lf,mn nd-dt-)& / Alwol dsic vle sSUi(oe>et i, oe>aa1 & / lo ny itnt aus iBTtlnd-rgt nd-dt+,mx; / Alwol dsic vle sSUi(oe>ih, oe>aa1 a) / lo ny itnt aus

}

/ Hle fnto ta alctsanwnd wt te * epr ucin ht loae e oe ih h gvndt adNL lf adrgtpitr.* ie aa n UL et n ih ones / src nd*nwoeitdt) tut oe eNd(n aa { src nd*nd =(tutnd* tut oe oe src oe) mlo(iefsrc nd); alcszo(tut oe)

nd-dt =dt; oe>aa aa nd-lf =NL; oe>et UL nd-rgt=NL; oe>ih UL } rtr(oe; eunnd)

/ Die pormt ts aoefntos/ * rvr rga o et bv ucin* itmi( n an) { src nd *ot=nwoe4; tut oe ro eNd() ro-lf ot>et =nwoe2; eNd() ro-rgt ot>ih =nwoe5; eNd() ro-lf-lf =nwoe1; ot>et>et eNd() ro-lf-rgt=nwoe3; ot>et>ih eNd() i(sS(ot) fiBTro) pit(I BT) rnf"s S"; es le pit(NtaBT) rnf"o S"; gthr) eca(; rtr 0 eun ;

}

Time Complexity: O(n) Auxiliary Space : O(1) if Function Call Stack size is not considered, otherwise O(n) METHOD 4(Using In-Order Traversal) Thanks to LJW489 for suggesting this method. 1) Do In-Order Traversal of the given tree and store the result in a temp array. 3) Check if the temp array is sorted in ascending order, if it is, then the tree is BST. Time Complexity: O(n) We can avoid the use of Auxiliary Array. While doing In-Order traversal, we can keep track of previously visited node. If the value of the currently visited node is less than the previous value, then tree is not BST. Thanks to ygos for this space optimization. bo iBTsrc nd*ro) ol sS(tut oe ot { sai src nd *rv=NL; ttc tut oe pe UL / taes tete i iodrfsinadke tako pe nd / rvre h re n nre aho n ep rc f rv oe i (ot f ro) { i (iBTro-lf) f !sS(ot>et) rtr fle eun as; / Alw ol dsic vle nds / los ny itnt aud oe i (rv! NL & ro-dt < pe-dt) f pe = UL & ot>aa = rv>aa

rtr fle eun as; pe =ro; rv ot } } rtr iBTro-rgt; eun sS(ot>ih)

rtr tu; eun re

The use of static variable can also be avoided by using reference to prev node as a parameter (Similar to this post). Sources: http://en.wikipedia.org/wiki/Binary_search_tree http://cslibrary.stanford.edu/110/BinaryTrees.html Please write comments if you find any bug in the above programs/algorithms or other ways to solve the same problem.

Related Topics:
Print a Binary Tree in Vertical Order | Set 2 (Hashmap based Method) Print Right View of a Binary Tree Red-Black Tree | Set 3 (Delete) Construct a tree from Inorder and Level order traversals Print all nodes at distance k from a given node Print a Binary Tree in Vertical Order | Set 1 Interval Tree Check if a given Binary Tree is height balanced like a Red-Black Tree
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Join the discussion… bakshi_993 •

4 days ago

can you please explain me the use of passing INT_MIN , INT_MAX in isBSTUtil
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