Leetcode:binary Tree postorder Traversal Problem Solving report

Binary Tree postorder Traversal

Given a binary tree, return the postorder traversal of its nodes ' values.

For example:
Given binary Tree {1,#,2,3},
1
\
2
/
3
return [3,2,1].

Note:recursive solution is trivial, could do it iteratively?

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Solution 1:

Recursive solution

1  PublicList<integer>postorderTraversal1 (TreeNode root) {2list<integer> ret =NewArraylist<integer>();3 Dfs (root, ret);4         returnret;5     }6     7     //Solution 1:rec8      Public voidDFS (TreeNode root, list<integer>ret) {9         if(Root = =NULL) {Ten             return; One         } A          - Dfs (Root.left, ret); - Dfs (root.right, ret); the Ret.add (root.val); -}

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Solution 2:

/**
* Sequential Traversal iterative method
* Http://www.youtube.com/watch?v=hv-mJUs5mvU
* http://blog.csdn.net/tang_jin2015/article/details/8545457
* Sequence from left to right is the same as the reverse order from right to left, so it's easy! Ha ha
* Flip with another stack
*/

1 //Solution 2:iterator2      PublicList<integer>postordertraversal (TreeNode root) {3list<integer> ret =NewArraylist<integer>();4         if(Root = =NULL) {5             returnret;6         }7         8stack<treenode> s =NewStack<treenode>();9Stack<integer> out =NewStack<integer>();Ten          One S.push (root); A          -          while(!S.isempty ()) { -TreeNode cur =S.pop (); the Out.push (cur.val); -              -             if(Cur.left! =NULL) { - S.push (cur.left); +             } -              +             if(Cur.right! =NULL) { A S.push (cur.right); at             } -         } -          -          while(!Out.isempty ()) { - Ret.add (Out.pop ()); -         } in          -         returnret; to}

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GITHUB:

Https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/PostorderTraversal.java

Leetcode:binary Tree postorder Traversal Problem Solving report