題目

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,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]
 

思路

題目大意

給定一個二叉樹，對樹進行層遍歷，但層遍歷方向會變化，開始從左到右，然後從右到左，不斷變化。

解題思路

方法一 樹的層遍歷 ，層的遍歷結果翻轉

對樹進行層遍歷，將特定層的遍歷結果進行翻轉操作。

方法二 stack 實現上層先遍歷的結點的子結點下層後遍歷

stack 實現上層先遍歷的結點的子結點下層後遍歷，從而 產生 本層從左到右遍歷 下層從右到左遍歷。

code

方法一 樹的層遍歷 ，層的遍歷結果翻轉

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 

class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        List<TreeNode> lst = new LinkedList<>();
        if(root==null) return res;
        int ilayer = 1;
        boolean
 
 isFromLeft;
        lst.add(root);
        while (!lst.isEmpty()){
            List<Integer> eleres = new ArrayList<>();
            int layerlen =  lst.size();
            for(int i = 0 ;i <layerlen ; i++){
                TreeNode tnode =  lst.get(0);
                lst.remove(0);
                eleres.add(tnode.val);
                if(tnode.left != null)   lst.add(tnode.left);
                if(tnode.right !=null) lst.add(tnode.right);   
            }
       
            if(ilayer%2==1){
                isFromLeft = true;
            }
            else {
                isFromLeft = false;
            }
            if(!isFromLeft){
                for(int i = 0 ,j = eleres.size() - 1;i<j;i++,j--){
                    Integer tmp = eleres.get(i);
                    eleres.set(i,eleres.get(j));
                    eleres.set(j,tmp);
                }
            }
            res.add(eleres);
            ilayer++;
        }
        return res;
    }
}

方法二 stack 實現上層先遍歷的結點的子結點下層後遍歷

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        if(root==null) return res;
        int ilayer = 1;
        boolean isFromLeft = true;
        stack.push(root);
        while (!stack.isEmpty()){
            List<Integer> eleres = new ArrayList<>();
            Stack<TreeNode> tstack = new Stack<>();
            while (!stack.isEmpty()){
                TreeNode tnode =  stack.pop();
                eleres.add(tnode.val);
                if(ilayer %2 == 1){
                    if(tnode.left != null)  tstack.push(tnode.left);
                    if(tnode.right != null) tstack.push(tnode.right);
                }
                else {
                    if(tnode.right != null) tstack.push(tnode.right);
                    if(tnode.left != null)  tstack.push(tnode.left);
                }
            }
            res.add(eleres);
            ilayer ++;
            stack = tstack;
        }
        return res;
    }
}