LeetCode刷题笔录 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]
]


哎,这题当时没刷过...

一开始想着一层用queue,一层用stack。发现这样不大对。从网上看到了一个有创意的解法:其实还是一个类似队列的操作,每次往队列里放也是先坐儿子后右儿子。只不过在奇数行的时候不插到队尾,而是往队列头插入。


public class Solution {
    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root == null)
            return result;
        search(root, 0, result);
        return result;
    }
    
    public void search(TreeNode node, int depth, ArrayList<ArrayList<Integer>> result){
        if(node == null)
            return;
        ArrayList<Integer> currentLevel = null;
        if(depth == result.size()){
            currentLevel = new ArrayList<Integer>();
            result.add(currentLevel);
        }
        else{
            currentLevel = result.get(depth);
        }
        //insert into the end of the list if the current depth is even
        //to the beginning of the list if the depth is odd
        if(depth % 2 == 0)
            currentLevel.add(node.val);
        else
            currentLevel.add(0, node.val);
        
        search(node.left, depth + 1, result);
        search(node.right, depth + 1, result);
    }
}

另外还可以用两个stack来进行traversal,具体的解法在 这里。感觉不用递归还是挺不错的

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