Leetcode 103.Binary tree Zigzag Level order traversal (binary tree Z-shape horizontal order) thinking and method of solving problems

Given a binary tree, return the zigzag level order traversal of its nodes ' values. (ie, from left-to-right, then right-to-left for the next level and alternate between).

For example:
Given binary Tree {3,9,20,#,#,15,7} ,

    3   /   9    /   7

Return its zigzag level order traversal as:

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

Train of thought: With two fork tree horizontal sequence problem-solving idea is similar, get the horizontal order result, then dual number chain list inversion can. ‘

The code is as follows:

/** * Definition for a binary tree node.  * public class TreeNode {* int val, * TreeNode left, * TreeNode right; * TreeNode (int x) {val = x;} *}    */public class Solution {list<list<integer>> List = new arraylist<list<integer>> ();        Public list<list<integer>> Zigzaglevelorder (TreeNode root) {DFS (0,root);        Exchange order every 1 lines for (int i = 1; i < list.size (); i = i+2) {list<integer> Al = List.get (i);        int len = Al.size ();        Reverse exchange for (int j = 0; J + J < Len-1; J + +) {int k = Al.get (j);        Al.set (J, Al.get (Len-1-j));        Al.set (Len-1-j, k);    }} return list; }/** * Sequence traversal, add list * @param the depth of the DEP tree based on depth * @param root root node */private void Dfs (int dep,treenode roo        T) {if (root = null) {return;        } list<integer> al;//Get Al value if (list.size () > Dep) {al = List.get (DEP) as appropriate;  }else{          Al = new arraylist<integer> ();        List.add (AL);        } dfs (Dep+1,root.left);        Al.add (Root.val);    DFS (Dep+1,root.right); }}

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Leetcode 103.Binary tree Zigzag Level order traversal (binary tree Z-shape horizontal order) thinking and method of solving problems