102. Binary Tree Level Order Traversal && 199. Binary Tree Right Side View-优快云博客

本文详细介绍了二叉树的层次遍历和右侧视图遍历两种算法。层次遍历从左到右按层级返回节点值，右侧视图遍历则返回从顶部到底部在右侧可见的节点值。

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102. Binary Tree Level Order Traversal

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,null,null,15,7],

return its level order traversal as:

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

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Tree Breadth-first Search

Hide Similar Problems

(M) Binary Tree Zigzag Level Order Traversal (E) Binary Tree Level Order Traversal II (E) Minimum Depth of Binary Tree (M) Binary Tree Vertical Order Traversal

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> returnVal = new ArrayList<List<Integer>>();
        if (root == null)
            return returnVal;
        Queue<TreeNode> nodes = new ArrayDeque<TreeNode>();
        nodes.offer(root);
        
        Queue<TreeNode> nextLevel = new ArrayDeque<TreeNode>();
        
        List<Integer> currentList = new ArrayList<Integer>();
        while(!nodes.isEmpty())
        {
            TreeNode next = nodes.poll();
            currentList.add(next.val);
            if(next.left != null)
                nextLevel.offer(next.left);
            if(next.right != null)
                nextLevel.offer(next.right);
            
            if(nodes.isEmpty())
            {
                nodes = nextLevel;
                nextLevel = new ArrayDeque<TreeNode>();
                returnVal.add(currentList);
                currentList = new ArrayList<Integer>();
            }
        }
        
        
        return returnVal;
    }
}

199. Binary Tree Right Side View

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

You should return [1, 3, 4].

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Tree Depth-first Search Breadth-first Search

Hide Similar Problems

(M) Populating Next Right Pointers in Each Node

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        Deque<TreeNode> level = new ArrayDeque<TreeNode>();
        List<Integer> results = new ArrayList<Integer>();
        if(root == null)
            return results;
        
        level.add(root);
        while(!level.isEmpty())
        {
            results.add(level.peekLast().val);
            
            Deque<TreeNode> currentLevel = new ArrayDeque<TreeNode>();
            while(!level.isEmpty())
            {
                TreeNode node = level.remove();
                if(node.left != null)
                    currentLevel.add(node.left);
                if(node.right != null)
                    currentLevel.add(node.right);
            }
            level = currentLevel;
        }
        
        return results;
    }
}

转载于:https://www.cnblogs.com/neweracoding/p/5229533.html