图的遍历总结

概念

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　　图的遍历有两种遍历方式：深度优先遍历(depth-first search)和广度优先遍历(breadth-first search)。

1.深度优先遍历

　　基本思路：首先从图中某个顶点V0出发，然后依次从V0相邻的顶点出发深度优先遍历，直至图中所有与V0路径相通的顶点都被访问了；若此时尚有顶点未被访问，则从中选一个顶点作为起始点，重复上述过程，直到所有的顶点都被访问。可以看出深度优先遍历是一个递归的过程。

    如下图中的一个无向图：

    

    其深度优先遍历得到的序列为：

    0->1->3->7->4->2->5->6

2.广度优先遍历

    基本思路：首先，从图的某个顶点V0之后，访问了V0之后，依次访问与V0相邻的未被访问的顶点，然后分别从这些顶点出发，广度优先遍历，直至所有的顶点都被访问完。也是递归的问题。

    如上面图中

    其广度优先遍历得到的序列：

    0->1->2->3->4->2->5->6->7

3.深度优先遍历和广度优先遍历区别

• 广度优先搜索与深度优先搜索的时间复杂度相同，只是遍历顺序不同。

深度优先搜索案例

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(Leetcode 173)
* Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
* Calling next() will return the next smallest number in the BST.
* Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

public class BSTIterator {

    private List<Integer> list = new ArrayList<Integer>();
    int cursor = -1;

    public static class TreeNode{
        int val;
        TreeNode left;
        TreeNode right;

        public TreeNode(int val) {
            this.val = val;
        }
    }

    /**
     * Your BSTIterator will be called like this:
     * BSTIterator i = new BSTIterator(root);
     * while (i.hasNext()) v[f()] = i.next();
     */

    public BSTIterator(TreeNode root) {
        helper(root);
    }
　　
    private void helper(TreeNode root) {
        if (root == null){
            return;
        }
        helper(root.left);
        list.add(root.val);
        helper(root.right);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if (cursor < list.size()-1) return true;
        else return false;
    }

    /** @return the next smallest number */
    public int next() {
        cursor++;
        return list.get(cursor);
    }

}

广度搜索举例

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Leetcode 199
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]
]
树节点的定义：

private class TreeNode{
        int val;
        TreeNode left;
        TreeNode right;

        public TreeNode(int val) {
            this.val = val;
        }
    }

广度优先搜索遍历:

public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        List<TreeNode> layer = new ArrayList<TreeNode>();

        if (root != null){
            layer.add(root);
        }

        while (!layer.isEmpty()){
            List<Integer> vals = new ArrayList<Integer>();
            List<TreeNode> temp = new ArrayList<TreeNode>();
            for (int i = 0; i < layer.size(); i++) {
                vals.add(layer.get(i).val);
                TreeNode left = layer.get(i).left;
                TreeNode right = layer.get(i).right;
                if (left != null){
                    temp.add(left);
                }
                if (right != null){
                    temp.add(right);
                }
            }
            result.add(vals);
            layer = temp;
        }

       return result;
    }