举例一个入口，利用一个Map构造HashMap时

/**     * Constructs a new HashMap with the same mappings as the     * specified Map.  The HashMap is created with     * default load factor (0.75) and an initial capacity sufficient to     * hold the mappings in the specified Map.     *     * @param   m the map whose mappings are to be placed in this map     * @throws  NullPointerException if the specified map is null     */    public HashMap(Map
     m) {        this.loadFactor = DEFAULT_LOAD_FACTOR;        putMapEntries(m, false);    }

然后就是调用putMapEntries方法，第二个参数其实可以看作细节，个人认为它和HashMap的子类LinkedHashMap有关，evict是逐出的意思，如果基于LinkedHashMap实现LRU缓存的话，这个evict参数正好就用上了。

/**     * Implements Map.putAll and Map constructor     *     * @param m the map     * @param evict false when initially constructing this map, else     * true (relayed to method afterNodeInsertion).     */    final void putMapEntries(Map
     m, boolean evict) {        int s = m.size();        if (s > 0) {            if (table == null) { // pre-size                float ft = ((float)s / loadFactor) + 1.0F;                int t = ((ft < (float)MAXIMUM_CAPACITY) ?                         (int)ft : MAXIMUM_CAPACITY);                if (t > threshold)                    threshold = tableSizeFor(t);            }            else if (s > threshold)                resize();            for (Map.Entry
     e : m.entrySet()) {                K key = e.getKey();                V value = e.getValue();                putVal(hash(key), key, value, false, evict);            }        }    }

可以看到在for循环中遍历旧的entrySet视图，然后将一个个的key-value对放入新构造的HashMap中，

for (Map.Entry
     e : m.entrySet()) {                K key = e.getKey();                V value = e.getValue();                putVal(hash(key), key, value, false, evict);            }

展开putVal(hash(key), key, value, false, evict);

/**     * Implements Map.put and related methods     *     * @param hash hash for key     * @param key the key     * @param value the value to put     * @param onlyIfAbsent if true, don't change existing value     * @param evict if false, the table is in creation mode.     * @return previous value, or null if none     */    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,                   boolean evict) {        Node
    
     [] tab; Node
     
       p; int n, i;        if ((tab = table) == null || (n = tab.length) == 0)            n = (tab = resize()).length;        if ((p = tab[i = (n - 1) & hash]) == null)            tab[i] = newNode(hash, key, value, null);        else {            Node
      
        e; K k;            if (p.hash == hash &&                ((k = p.key) == key || (key != null && key.equals(k))))                e = p;            else if (p instanceof TreeNode)                e = ((TreeNode
       
        )p).putTreeVal(this, tab, hash, key, value);            else {                for (int binCount = 0; ; ++binCount) {                    if ((e = p.next) == null) {                        p.next = newNode(hash, key, value, null);                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st                            treeifyBin(tab, hash);                        break;                    }                    if (e.hash == hash &&                        ((k = e.key) == key || (key != null && key.equals(k))))                        break;                    p = e;                }            }            if (e != null) { // existing mapping for key                V oldValue = e.value;                if (!onlyIfAbsent || oldValue == null)                    e.value = value;                afterNodeAccess(e);                return oldValue;            }        }        ++modCount;        if (++size > threshold)            resize();        afterNodeInsertion(evict);        return null;    }
       
      
     
    

通过hash(key)定位到HashMap中tab数组的索引，如果这个数组元素的头节点正好是TreeNode类型，那么就将执行

e = ((TreeNode
    
     )p).putTreeVal(this, tab, hash, key, value);
    

此时this是HashMap自身。

putTreeVal考虑两大情况，1）key已经存在这个红黑树中当中了，就直接放回对应的那个节点；2）从红黑树的root节点开始遍历，定位到要插入的叶子节点，插入新节点；putTreeVal除了要维护红黑树的平衡外（可以参考TreeMap源码），还需要维护节点之间的前后关系，这里似乎同时是在维护双向链表关系。

/**         * Tree version of putVal.         */        final TreeNode
    
      putTreeVal(HashMap
     
       map, Node
      
       [] tab,                                       int h, K k, V v) {            Class
        kc = null;            boolean searched = false;            TreeNode
       
         root = (parent != null) ? root() : this;            for (TreeNode
        
          p = root;;) {                int dir, ph; K pk;                if ((ph = p.hash) > h)                    dir = -1;                else if (ph < h)                    dir = 1;                else if ((pk = p.key) == k || (k != null && k.equals(pk)))                    return p;                else if ((kc == null &&                          (kc = comparableClassFor(k)) == null) ||                         (dir = compareComparables(kc, k, pk)) == 0) {                    if (!searched) {                        TreeNode
         
           q, ch; searched = true; if (((ch = p.left) != null && (q = ch.find(h, k, kc)) != null) || ((ch = p.right) != null && (q = ch.find(h, k, kc)) != null)) return q; } dir = tieBreakOrder(k, pk); } TreeNode
          
            xp = p; if ((p = (dir <= 0) ? p.left : p.right) == null) { Node
           
             xpn = xp.next; TreeNode
            
              x = map.newTreeNode(h, k, v, xpn); if (dir <= 0) xp.left = x; else xp.right = x; xp.next = x; x.parent = x.prev = xp; if (xpn != null) ((TreeNode
             
              )xpn).prev = x; moveRootToFront(tab, balanceInsertion(root, x)); return null; } } }
             
            
           
          
         
        
       
      
     
    

下面重点分析putTreeVal方法

1 首先找到root节点，

TreeNode
    
      root = (parent != null) ? root() : this;
    

这里的this是指TreeNode自己，从某个节点一直往上溯，直到parent==null的情况

2 递归遍历root判断节点之间的hash大小，如果hash值相等采用key比较等然后采用左子树或者右子树，继续遍历（关于key值大小的比较算是细节的地方，这里暂且代入String类型的key解读源码以图整体思路流畅）3 如果遍历到了叶子节点比如上一步采用左子树，而左子树刚好是叶子节点，p == null此时递归遍历结束

if ((p = (dir <= 0) ? p.left : p.right) == null) {                    Node
    
      xpn = xp.next;                    TreeNode
     
       x = map.newTreeNode(h, k, v, xpn);                    if (dir <= 0)                        xp.left = x;                    else                        xp.right = x;                    xp.next = x;                    x.parent = x.prev = xp;                    if (xpn != null)                        ((TreeNode
      
       )xpn).prev = x;                    moveRootToFront(tab, balanceInsertion(root, x));                    return null;                }
      
     
    

xp是叶子节点的父节点，这个节点不是null，叶子节点p一定是null

新增一个节点x，next指向原来父节点的.next，x就是新增的叶子节点1) 处理红黑树的关系父节点xp和叶子节点x的关系，落在左子树还是右子树；x的parent指向父节点xp x.parent = xp最后保持红黑树平衡2）处理双向链表的关系类似于在xp-->xpn(xp.next)中间插入新的节点x，即

x = map.newTreeNode(h, k, v, xpn);x.prev = xp//如果xpn不是null，则处理xpn的prev((TreeNode
    
     )xpn).prev = x;
    

图示

3) 保持红黑树平衡

moveRootToFront(tab, balanceInsertion(root, x));