Posts Tagged 【bfs】Binary Tree Level Order Traversal I && II

Binary Tree Level Order Traversal

 Total Accepted: 47754 Total Submissions: 161177My Submissions

Question Solution 

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

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

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/*【BFS通常框架】通常使用队列实现FIFO    初始化队列Q    Q={起点s}，s标记为已经访问    while(Q不为空) {        取队首head，head出队        if(head处达到目标) {...}        所有head未访问的相邻节点加入队列        标记head已经被访问    }eg：int[] queue = new int[n];int head = 0;tail = 0;queue[tail++] = root;while(head < tail) {   //do thing }【三种思路，最后一种最为简洁】===【每个节点记录自己所在的层数】一个list记录遍历的先后结构，一个list记录遍历的节点的层数如果输出遍历路线，一般还要保持父节点的遍历中存的位置最后将两个list的结果对照来输出===【用一个符号来比较层的转移】用一个变量节点来记录下一层的开始节点while(!Q.isEmpty()) {    Node n = Q.peek();    Node n2depth = null'    while(n != n2depth && !Q.isEmpty()) {        Q.remove();        if(n2depth == null) {            n2depth = n not null 子节点        }        Q.add(n not null 子节); n = Q.peek();    }    //do thing for this layer}===【用两个队列来存上下两层的数据】从root开始，放到第一个队列接着取出第一队列，将其相邻或子节点放到下一个队列如此反复*//** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */  /*Input:{}Output:nullExpected:[] */public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> lists = new ArrayList<List<Integer>>();        if(root == null) return lists;        List<Integer> list; //注意每次都要new        Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.add(root);        while(!queue.isEmpty()) {            Queue<TreeNode> temp = new LinkedList<TreeNode>();            list = new ArrayList<Integer>();            while(!queue.isEmpty()) {                TreeNode n = queue.remove();                list.add(n.val);                if(n.left != null) temp.add(n.left);                if(n.right != null) temp.add(n.right);            }            lists.add(list);            queue = temp;        }        return lists;    }}

Binary Tree Level Order Traversal II

 Total Accepted: 38751 Total Submissions: 125221My Submissions

Question Solution 

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */  /*没啥好搞的 Colletcions.reverse(lists); */public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        List<List<Integer>> lists = new ArrayList<List<Integer>>();        if(root == null) return lists;        List<Integer> list; //注意每次都要new        Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.add(root);        while(!queue.isEmpty()) {            Queue<TreeNode> temp = new LinkedList<TreeNode>();            list = new ArrayList<Integer>();            while(!queue.isEmpty()) {                TreeNode n = queue.remove();                list.add(n.val);                if(n.left != null) temp.add(n.left);                if(n.right != null) temp.add(n.right);            }            lists.add(list);            queue = temp;        }        Collections.reverse(lists);        return lists;    }}

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