举例一个入口，利用一个Map构造HashMap时
/**
* Constructs a new <tt>HashMap</tt> with the same mappings as the
* specified <tt>Map</tt>. The <tt>HashMap</tt> is created with
* default load factor (0.75) and an initial capacity sufficient to
* hold the mappings in the specified <tt>Map</tt>.
*
* @param m the map whose mappings are to be placed in this map
* @throws NullPointerException if the specified map is null
*/
public HashMap(Map<? extends K, ? extends V> 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<? extends K, ? extends V> 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<? extends K, ? extends V> 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<? extends K, ? extends V> 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<K,V>[] tab; Node<K,V> 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<K,V> 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<K,V>)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<K,V>)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<K,V> putTreeVal(HashMap<K,V> map, Node<K,V>[] tab,
int h, K k, V v) {
Class<?> kc = null;
boolean searched = false;
TreeNode<K,V> root = (parent != null) ? root() : this;
for (TreeNode<K,V> 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<K,V> 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<K,V> xp = p;
if ((p = (dir <= 0) ? p.left : p.right) == null) {
Node<K,V> xpn = xp.next;
TreeNode<K,V> 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<K,V>)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<K,V> 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<K,V> xpn = xp.next;
TreeNode<K,V> 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<K,V>)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<K,V>)xpn).prev = x;图示
3) 保持红黑树平衡
moveRootToFront(tab, balanceInsertion(root, x));