[数据结构与算法]【面试算法题总结04】广度优先搜索

广度优先搜索：

每次将一个节点周围的所有节点加入队列。适合找最短距离，但空间复杂度高，而且没有深度优先搜索的递归代码好写。
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例题1：二叉树的层序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if(root==null){
            return new ArrayList<>();
        }
        List<List<Integer>> result=new ArrayList<>();
        Queue<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        while(queue.size()!=0){
            int size=queue.size();
            List<Integer> temp=new ArrayList<>();
            for(int i=0;i<size;++i){
                TreeNode currentNode=queue.poll();
                temp.add(currentNode.val);
                if(currentNode.left!=null){
                    queue.offer(currentNode.left);
                }
                if(currentNode.right!=null){
                    queue.offer(currentNode.right);
                }               
            }
            result.add(temp);
        }
        return result;
    }
}

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例题2：二叉树的层序遍历 II

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        if(root==null){
            return new ArrayList<>();
        }
        List<List<Integer>> result=new ArrayList<>();
        Queue<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        while(queue.size()!=0){
            int size=queue.size();
            List<Integer> temp=new ArrayList<>();
            for(int i=0;i<size;++i){
                TreeNode currentNode=queue.poll();
                temp.add(currentNode.val);
                if(currentNode.left!=null){
                    queue.offer(currentNode.left);
                }
                if(currentNode.right!=null){
                    queue.offer(currentNode.right);
                }               
            }
            result.add(0,temp);
        }
        return result;
    }
}

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例题3：从上到下打印二叉树 III

/**
 * 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>> levelOrder(TreeNode root) {
        if(root==null){
            return new ArrayList<>();
        }
        List<List<Integer>> result=new ArrayList<>();
        Queue<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        int depth=1;
        while(queue.size()!=0){
            int size=queue.size();
            List<Integer> temp=new ArrayList<>();
            for(int i=0;i<size;++i){
                TreeNode currentNode=queue.poll();
                temp.add(currentNode.val);
                if(currentNode.left!=null){
                    queue.offer(currentNode.left);
                }
                if(currentNode.right!=null){
                    queue.offer(currentNode.right);
                }               
            }
            if(depth%2==0){
                Collections.reverse(temp);
            }
            result.add(temp);
            ++depth;
        }
        return result;
    }
}

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例题4：二叉树的最小深度

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if(root==null){
            return 0;
        }
        Queue<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        int depth=1;
        while(queue.size()!=0){
            int size=queue.size();
            for(int i=0;i<size;++i){
                TreeNode currentNode=queue.poll();
                if(currentNode.left==null&&currentNode.right==null){
                    return depth;
                }
                if(currentNode.left!=null){
                    queue.offer(currentNode.left);
                }
                if(currentNode.right!=null){
                    queue.offer(currentNode.right);
                }               
            }
            ++depth;
        }
        return depth;
    }
}

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