37. Binary Tree zigzag level order traversal & Binary Tree inorder Traversal

Binary Tree zigzag level order traversal

Given a binary tree, returnZigzag level orderTraversal of its nodes 'values. (ie, from left to right, then right to left for the next level and alternate ).

For example: given binary tree{3, 9, 20, #, #, 15, 7},

 
3/9 20/15 7
 

 

Return its zigzag level order traversal:

 
[[3], [20, 9], [15, 7]
 

Idea: Use two queues (one can be read sequentially, so we use vector to simulate it), and each queue has a layer of nodes.

/*** Definition for binary tree * struct treenode {* int val; * treenode * left; * treenode * right; * treenode (int x): Val (x ), left (null), right (null) {}*}; */void getque1 (vector <treenode *> & Q1, queue <treenode *> & Q2, vector <vector <int> & VEC) {While (! Q2.empty () {treenode * P = q2.front (); q2.pop (); If (p-> left) q1.push _ back (p-> left ); if (p-> right) q1.push _ back (p-> right);} If (q1.size () = 0) return; vector <int> vec2; for (INT I = q1.size ()-1; I> = 0; -- I) vec2.push _ back (Q1 [I]-> Val); Vec. push_back (vec2);} void getque2 (queue <treenode *> & Q2, vector <treenode *> & Q1, vector <int> & VEC) {If (q1.size () = 0) return; vector <int> vec2; For (INT I = 0; I <q1.size (); ++ I) {If (Q1 [I]-> left) {q2.push (Q1 [I]-> left ); vec2.push _ back (Q1 [I]-> left-> Val);} If (Q1 [I]-> right) {q2.push (Q1 [I]-> right ); vec2.push _ back (Q1 [I]-> right-> Val) ;}} if (vec2.size () Vec. push_back (vec2); q1.clear ();} class solution {public: vector <int> zigzaglevelorder (treenode * root) {vector <int> VEC; if (root = NULL) return VEC; queue <treenode *> Q2; vector <treenod E *> q1; q2.push (Root); Vec. push_back (vector <int> (1, root-> Val); While (! Q2.empty () {getque1 (Q1, q2, VEC); getque2 (Q2, Q1, VEC) ;}return VEC ;}};

 

Binary Tree inorder Traversal

OJ: https://oj.leetcode.com/problems/binary-tree-inorder-traversal/

Given a binary tree, returnInorderTraversal of its nodes 'values.

For example: given binary tree{1, #, 2, 3},

 
1 2/3
 

 

Return[1, 3, 2].

Note:Recursive solution is trivial, cocould You Do It iteratively?

Question: two methods: 1. STACK: O (n) time, O (n) space. 2. Morris traversal (construct the clue tree), O (n) time, O (1) space.

1. Use stacks

/*** Definition for binary tree * struct treenode {* int val; * treenode * left; * treenode * right; * treenode (int x): Val (x ), left (null), right (null) {}*}; */class solution {public: vector <int> inordertraversal (treenode * root) {vector <int> VEC; if (root = NULL) return VEC; treenode * P = root; stack <treenode *> st; ST. push (p); While (p-> left) {P = p-> left; ST. push (p) ;}while (! St. empty () {treenode * q = ST. top (); ST. pop (); Vec. push_back (Q-> Val); If (Q-> right) {q = Q-> right; ST. push (Q); While (Q-> left) {q = Q-> left; ST. push (q) ;}}return VEC ;}};
 

2. Morris Traversal

/*** Definition for binary tree * struct treenode {* int val; * treenode * left; * treenode * right; * treenode (int x): Val (x ), left (null), right (null) {}*}; */class solution {public: vector <int> inordertraversal (treenode * root) {vector <int> VEC; treenode * cur, * pre; cur = root; while (cur) {If (cur-> left = NULL) {Vec. push_back (cur-> Val); cur = cur-> right;} else {pre = cur-> left; while (pre-> Right & pr E-> right! = Cur) Pre = pre-> right; if (pre-> right = NULL) {pre-> right = cur; cur = cur-> left ;} else {pre-> right = NULL; Vec. push_back (cur-> Val); cur = cur-> right ;}}return VEC ;}};
 

 

37. Binary Tree zigzag level order traversal & Binary Tree inorder Traversal