"Leetcode-Interview algorithm classic-java implementation" "107-binary Tree Level Order Traversal II (binary sequence Traversal II)"

"107-binary Tree Level Order Traversal II (binary tree Sequence Traversal II)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" Original Question

Given a binary tree, return the bottom-up level order traversal of its nodes ' values. (ie, from the left-to-right, the level by level from the leaf to root).
For example:
Given binary Tree {3,9,20,#,#,15,7},

    3     9  20    /     15   7

Return its bottom-up level order traversal as:

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

Main Topic

Given a binary tree, the sequence traversal is done from the bottom down.

Thinking of solving problems

The tree is sequentially traversed, and the results of each layer are placed on the head of the resulting list.

Code Implementation

publicclass TreeNode {    int val;    TreeNode left;    TreeNode right;    TreeNode(int x) { val = x; }}

Algorithm implementation class

ImportJava.Util.*; PublicClass Solution { Public List<List<Integer>>Levelorderbottom (TreeNode root) {List<List<Integer>> List = NewLinkedList<>();if(Root== NULL) {return List; } Deque<TreeNode>Cur= NewLinkedList<>(); Deque<TreeNode>Nxt= NewLinkedList<>(); Deque<TreeNode>Exc= NewLinkedList<>();        TreeNode tmp; Cur.Add (root); while(!Cur.IsEmpty ()) {List<Integer>Layout= NewArrayList<>(); while(!Cur.IsEmpty ()) {tmp=Cur.Remove ();if(TMP.Left!= NULL) {NXT.Add (tmp.left); }if(TMP.Right!= NULL) {NXT.Add (tmp.right); } layout.Add (tmp.Val); } exc=Cur Cur=Nxt Nxt=ExcList.Add0, layout); }return List; }}

Evaluation Results

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"Leetcode-Interview algorithm classic-java implementation" "107-binary Tree Level Order Traversal II (binary sequence Traversal II)"