Binary Tree Level Order Traversal II

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Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

对树的逆层次遍历。我们不可能从底部往上走，所以我们至顶向下走，不断的在链表前端插入新的一层节点。

public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        int cnt = 0;int f = 0;int t = 1;int index; List<List<Integer>> lsts = new ArrayList<List<Integer>>();List<Integer> lst = new ArrayList<Integer>();List<TreeNode> nodeList = new ArrayList<TreeNode>();nodeList.add(root);if(root == null){return lsts;}index = 0;while(t != 0){            if(f == t){            t = cnt;            f = 0;            cnt = 0;            lsts.add(0,lst);            lst = new ArrayList<Integer>();            }            else{            TreeNode node = nodeList.get(index ++);            if(node != null){            f ++;            lst.add(node.val);            }            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            }}    return lsts;    }}

然而只占胜了23.52%，于是我们选择还是按顺序放入链表，在进行反转。

public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        int cnt = 0;int f = 0;int t = 1;int index; List<List<Integer>> lsts = new ArrayList<List<Integer>>();List<Integer> lst = new ArrayList<Integer>();List<TreeNode> nodeList = new ArrayList<TreeNode>();nodeList.add(root);if(root == null){return lsts;}index = 0;while(t != 0){            if(f == t){            t = cnt;            f = 0;            cnt = 0;            lsts.add(lst);            lst = new ArrayList<Integer>();            }            else{            TreeNode node = nodeList.get(index ++);            if(node != null){            f ++;            lst.add(node.val);            }            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            }}int from = 0,to = lsts.size() - 1;while(from < to){List<Integer> tempF = lsts.get(from);List<Integer> tempE = lsts.get(to);lsts.set(from, tempE);lsts.set(to, tempF);from ++;to --;}    return lsts;    }}

此刻战胜了69.49% 。当然了，自然是可以利用递归解决的，递归也可以解决的。

        public List<List<Integer>> levelOrderBottom(TreeNode root) {            List<List<Integer>> wrapList = new LinkedList<List<Integer>>();            levelMaker(wrapList, root, 0);            return wrapList;        }                public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {            if(root == null) return;            if(level >= list.size()) {                list.add(0, new LinkedList<Integer>());            }            levelMaker(list, root.left, level+1);            levelMaker(list, root.right, level+1);            list.get(list.size()-level-1).add(root.val);        }

这个方案战胜了69.49%

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