LeetCode 49 Binary Tree Zigzag Level Order Traversal

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  20
   / \
  15  7

return its zigzag level order traversal as:

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

分析：

层序遍历的变种，每层反向，可用栈代替队列来实现反向。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(root==null) return result;
        Stack<TreeNode> st1 = new Stack<TreeNode>();
        Stack<TreeNode> st2 = new Stack<TreeNode>();
        st1.push(root);
        while(st1.size()>0 || st2.size()>0 ){
            List<Integer> level = new ArrayList<Integer>();
            if(st1.size()>0){
                while(st1.size()>0){
                    TreeNode temp = st1.pop();
                    level.add(temp.val);
                    if(temp.left != null) st2.push(temp.left);
                    if(temp.right != null) st2.push(temp.right);
                }
                result.add(level);
                continue;
            }
            if(st2.size()>0){
                while(st2.size()>0){
                    TreeNode temp = st2.pop();
                    level.add(temp.val);
                    if(temp.right!= null) st1.push(temp.right);
                    if(temp.left != null) st1.push(temp.left);
                }
                result.add(level);
            }
        }
        return result;
    }
}