node.sort()

node.sort(compare)

Sorts the children of this node, if any, and each of this node’s descendants’ children, in pre-order traversal using the specified compare function, and returns this node. The specified function is passed two nodes a and b to compare. If a should be before b, the function must return a value less than zero; if b should be before a, the function must return a value greater than zero; otherwise, the relative order of a and b are not specified. See array.sort for more.

Unlike node.sum, the compare function is passed two nodes rather than two nodes’ data. For example, if the data has a value property, this sorts nodes by the descending aggregate value of the node and all its descendants, as is recommended for circle-packing:

root
    .sum(function(d) { return d.value; })
    .sort(function(a, b) { return b.value - a.value; });

Similarly, to sort nodes by descending height (greatest distance from any descendant leaf) and then descending value, as is recommended for treemaps and icicles:

root
    .sum(function(d) { return d.value; })
    .sort(function(a, b) { return b.height - a.height || b.value - a.value; });

To sort nodes by descending height and then ascending id, as is recommended for trees and dendrograms:

root
    .sum(function(d) { return d.value; })
    .sort(function(a, b) { return b.height - a.height || a.id.localeCompare(b.id); });

You must call node.sort before invoking a hierarchical layout if you want the new sort order to affect the layout; see node.sum for an example.